【数据结构】二叉树的遍历：前序，中序，后序的递归结构遍历2 A( g! M9 j- V# |
5 ~; F( v( S7 Q4 J[color=rgba(0, 0, 0, 0.749019607843137)]文章目录4 C7 }9 r% ?& Q0 n' n; ?0 d
[color=rgba(0, 0, 0, 0.749019607843137)]前言
# c5 X" o/ l+ `* ~5 U: ^$ [5 M[color=rgba(0, 0, 0, 0.749019607843137)]1.二叉树的遍历方式
: X% S) C( D- v$ N" i[color=rgba(0, 0, 0, 0.749019607843137)]2.二叉树的遍历及相关函数（代码实现）4 e# u/ w# T1 m" G2 l& Z( ], P
[color=rgba(0, 0, 0, 0.749019607843137)]2.1前序/中序/后序的遍历" G2 E! v$ R+ i6 m- N6 z
[color=rgba(0, 0, 0, 0.749019607843137)]2.2计算二叉树的大小+ {' f; ^$ h" i% Z' p8 e2 m
[color=rgba(0, 0, 0, 0.749019607843137)]2.3计算二叉树叶子结点的个数2 V# f* @1 }0 N2 U2 B1 E$ A( \( B" P
[color=rgba(0, 0, 0, 0.749019607843137)]2.4计算二叉树的高度 P# T: P$ |& V# S% ?" f
[color=rgba(0, 0, 0, 0.749019607843137)]2.5计算第K层结点的个数
4 K' ~4 O+ W) x* N( j) d$ h9 @[color=rgba(0, 0, 0, 0.749019607843137)]2.6二叉树查找" F* b p: `- v1 V. j/ M) L
[color=rgba(0, 0, 0, 0.749019607843137)]前言
4 t$ A8 g- |# m; ^[color=rgba(0, 0, 0, 0.749019607843137)]在学习二叉树的遍历之前，我们需要先创建一棵二叉树，然后才能学习其相关的基本操作，由于现在我们对二叉树结构的掌握还在初阶部分，为了降低大家的学习成本，我们先手动快速创建一棵简单的二叉树，快速进入二叉树的操作学习，这个方法在我们调试程序代码的时候，也非常适用。等二叉树结构了解的差不多时，我们再继续研究二叉树真正的创建方式。
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n `# H7 M0 y& ~" C" g t [[color=rgba(0, 0, 0, 0.749019607843137)]1.二叉树的遍历方式
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5 b$ D% A. k" I+ f' J- t2 a; V. R[color=rgba(0, 0, 0, 0.749019607843137)]按照规则，二叉树的遍历有：前序/中序/后序的递归结构遍历访问顺序：# _5 ~. g# |( p! _1 V8 `
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6 {" [6 n# J/ A[color=rgba(0, 0, 0, 0.749019607843137)]1. 前序遍历（先序，先根):根——左子树——右子树
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[color=rgba(0, 0, 0, 0.749019607843137)]2. 中序遍历（中根）：左子树——根——右子树
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[color=rgba(0, 0, 0, 0.749019607843137)]3. 后序遍历（后根）：左子树——右子树——根 l9 y. a8 M$ i2 ~2 e- l% l, B+ Z; y
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5 b1 F/ U; y. j1 Z; b8 O. R[color=rgba(0, 0, 0, 0.749019607843137)]2.二叉树的遍历及相关函数（代码实现）
3 N& j3 _& F0 {& V2 y+ a9 ^[color=rgba(0, 0, 0, 0.749019607843137)]思路：分而治之
$ Z5 W+ g; ~( I u[color=rgba(0, 0, 0, 0.749019607843137)]1.首先我们要用简单的方式先创建出一棵二叉树，并赋予数据；
z1 t6 Q5 o& C9 q1 X" Q9 i( c* B/ w[color=rgba(0, 0, 0, 0.749019607843137)]2.采用递归的方式，分别实现前序/中序/后序遍历这棵二叉树；
@4 g. ^. g" K2 ^4 ~8 y, J, `7 b% _[color=rgba(0, 0, 0, 0.749019607843137)]3.尝试计算这个二叉树的大小（利用递归）；6 u2 t, \, K' c
[color=rgba(0, 0, 0, 0.749019607843137)]4.尝试计算叶子结点的个数（利用递归）；$ W% ?6 X5 c& \2 M8 K' W6 c% G
[color=rgba(0, 0, 0, 0.749019607843137)]5.尝试计算二叉树的高度（利用递归）；6 o% S. {/ l+ l9 C- o5 u9 G
[color=rgba(0, 0, 0, 0.749019607843137)]6.尝试写出计算第K层结点的个数的函数（利用递归）；: |$ Y" q& U1 j/ F0 w! ]8 @
[color=rgba(0, 0, 0, 0.749019607843137)]7.尝试写出二叉树查找的函数（利用递归）。
$ \# e: ?* i" [+ a: w[color=rgba(0, 0, 0, 0.749019607843137)]
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[color=rgba(0, 0, 0, 0.749019607843137)]2.1前序/中序/后序的遍历
9 _" b% A* b5 u/ u# g; s2 R[color=rgba(0, 0, 0, 0.749019607843137)]#define _CRT_SECURE_NO_WARNINGS 11 s5 E# @5 v' ~) s
[color=rgba(0, 0, 0, 0.749019607843137)]#include<stdio.h>
& @3 ?6 a0 h7 `# d[color=rgba(0, 0, 0, 0.749019607843137)]#include<assert.h>
/ h) F. r8 E# @2 w- m[color=rgba(0, 0, 0, 0.749019607843137)]#include<stdlib.h>0 t9 s* Y! E. L* _/ P4 T8 X1 o
[color=rgba(0, 0, 0, 0.749019607843137)]) N" c! C0 p! {5 @) O3 S
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[color=rgba(0, 0, 0, 0.749019607843137)]typedef int BTDataType;- L" P- Y! C$ y" a" R1 J5 Z
[color=rgba(0, 0, 0, 0.749019607843137)]& z8 u" F% P$ W {
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[color=rgba(0, 0, 0, 0.749019607843137)]//定义二叉树结点的结构体5 ^4 C1 h6 ~# z# H' Z
[color=rgba(0, 0, 0, 0.749019607843137)]typedef struct BinaryTreeNode
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. i" ]' U& D; |" D/ W[color=rgba(0, 0, 0, 0.749019607843137)] BTDataType data;
! n- c$ I6 Y7 w+ L[color=rgba(0, 0, 0, 0.749019607843137)] struct BinaryTreeNode* left;
! i4 }2 Z- Q P# {7 G2 S[color=rgba(0, 0, 0, 0.749019607843137)] struct BinaryTreeNode* right;9 L5 V0 k7 \ H0 w4 M
[color=rgba(0, 0, 0, 0.749019607843137)]}BTNode;9 X1 X. v# K4 W5 M' ^$ c# |
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6 h2 D9 J. ~- V# w1 V[color=rgba(0, 0, 0, 0.749019607843137)]//前序遍历6 r9 A# ?/ a& x) ^. F5 \1 A+ q8 b! B
[color=rgba(0, 0, 0, 0.749019607843137)]void PreOrder(BTNode* root)$ z, H( y) d2 v( e5 ^' w
[color=rgba(0, 0, 0, 0.749019607843137)]{! X8 J' b* Y% ] x
[color=rgba(0, 0, 0, 0.749019607843137)] if (root == NULL)7 Q$ L. @8 s- B) F2 |8 J% |. J
[color=rgba(0, 0, 0, 0.749019607843137)] {
/ T) ~5 s( P* J! b[color=rgba(0, 0, 0, 0.749019607843137)] printf("NULL ");
Y/ s2 \# i# f8 @ e$ z: u[color=rgba(0, 0, 0, 0.749019607843137)] return;
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[color=rgba(0, 0, 0, 0.749019607843137)] printf("%d ", root->data);
5 E2 h3 E2 O' J0 r: p) |[color=rgba(0, 0, 0, 0.749019607843137)] PreOrder(root->left);
6 Y4 l+ i. }: x[color=rgba(0, 0, 0, 0.749019607843137)] PreOrder(root->right);/ X' @, E3 Y, t* _- r
[color=rgba(0, 0, 0, 0.749019607843137)]}
- a+ E4 |- w3 ][color=rgba(0, 0, 0, 0.749019607843137)]//中序遍历
7 Q+ X- k+ _* B8 q[color=rgba(0, 0, 0, 0.749019607843137)]void InOrder(BTNode* root)% j; x a" @, U
[color=rgba(0, 0, 0, 0.749019607843137)]{
2 }5 ~$ J, O# `8 P" q6 B) p4 y[color=rgba(0, 0, 0, 0.749019607843137)] if (root == NULL)) s% o" X$ |, w! f: ?+ t
[color=rgba(0, 0, 0, 0.749019607843137)] {9 [3 N/ ~: `: M$ ]4 j. ]/ Z
[color=rgba(0, 0, 0, 0.749019607843137)] printf("NULL ");4 ^" u5 J, t: a6 ^
[color=rgba(0, 0, 0, 0.749019607843137)] return;- c* B# ^* l: s7 C$ C( t
[color=rgba(0, 0, 0, 0.749019607843137)] }* Y: V, L! D1 M( ^ i' b4 V
[color=rgba(0, 0, 0, 0.749019607843137)] InOrder(root->left);
( c! w: F4 f% p5 T, p; W[color=rgba(0, 0, 0, 0.749019607843137)] printf("%d ", root->data);2 c8 g C e9 D6 O c! C- n+ e+ I' d
[color=rgba(0, 0, 0, 0.749019607843137)] InOrder(root->right);
4 {2 S9 K. W7 h0 F[color=rgba(0, 0, 0, 0.749019607843137)]}
3 c' b+ i! f% A, u, ], Y; V6 ?$ W1 T[color=rgba(0, 0, 0, 0.749019607843137)]//后序遍历
: _. v5 F9 B5 M! ?[color=rgba(0, 0, 0, 0.749019607843137)]void PostOrder(BTNode* root)
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/ ]- g+ g. N* [' R[color=rgba(0, 0, 0, 0.749019607843137)] if (root == NULL)
" X1 x' h+ e# k[color=rgba(0, 0, 0, 0.749019607843137)] {
5 M4 A! Q3 F: T8 r[color=rgba(0, 0, 0, 0.749019607843137)] printf("NULL ");
1 \3 v% m1 ]) g3 ], g" K$ G% d. D( c2 J[color=rgba(0, 0, 0, 0.749019607843137)] return;
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, l* B( n5 b+ v* T[color=rgba(0, 0, 0, 0.749019607843137)] PostOrder(root->left);" u$ x% r4 v2 |% ]; n
[color=rgba(0, 0, 0, 0.749019607843137)] PostOrder(root->right);) K: u( E+ w0 E* n9 J% [4 A6 J( I
[color=rgba(0, 0, 0, 0.749019607843137)] printf("%d ", root->data);
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[color=rgba(0, 0, 0, 0.749019607843137)]//先创建一个简单的二叉树结构" H: [/ f1 G# W' R
[color=rgba(0, 0, 0, 0.749019607843137)]BTNode* CreateTree(); L' c, V0 f* v- h7 J
[color=rgba(0, 0, 0, 0.749019607843137)]{4 S5 M8 K! |0 _0 g8 S: z9 D
[color=rgba(0, 0, 0, 0.749019607843137)] //先动态开辟6个结点的空间
* K* p# ?/ q' z[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* n1 = (BTNode*)malloc(sizeof(BTNode));
W3 r8 _: f2 l1 M[color=rgba(0, 0, 0, 0.749019607843137)] assert(n1);2 z: d, E: k8 H8 p/ B
[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* n2 = (BTNode*)malloc(sizeof(BTNode));
' a0 Z+ q) @7 P# _& T: _* P[color=rgba(0, 0, 0, 0.749019607843137)] assert(n2);
' E$ b* t q; u. U& s# @[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* n3 = (BTNode*)malloc(sizeof(BTNode));# ` n' Q* u }2 [+ b
[color=rgba(0, 0, 0, 0.749019607843137)] assert(n3);
' T, \1 N) K& `% Y[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* n4 = (BTNode*)malloc(sizeof(BTNode));4 u6 F* f# r& K: A |2 j
[color=rgba(0, 0, 0, 0.749019607843137)] assert(n4);
. E E' n# |. q* z8 p[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* n5 = (BTNode*)malloc(sizeof(BTNode)); X- A K& b# F! @
[color=rgba(0, 0, 0, 0.749019607843137)] assert(n5);
3 {. S1 e( G: K0 X0 J8 N) K[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* n6 = (BTNode*)malloc(sizeof(BTNode));; n Y- ]" T: g5 ~5 s
[color=rgba(0, 0, 0, 0.749019607843137)] assert(n6);$ B+ q9 `" o. i' d7 L7 [
[color=rgba(0, 0, 0, 0.749019607843137)]# H9 R3 \. q. ~4 N6 X+ \
$ T+ P4 c$ [9 E6 U[color=rgba(0, 0, 0, 0.749019607843137)] n1->data = 1;6 n) M3 m4 o5 N+ r$ r2 v2 T$ E
[color=rgba(0, 0, 0, 0.749019607843137)] n2->data = 2;
7 B* j( m4 Y( _ y0 f[color=rgba(0, 0, 0, 0.749019607843137)] n3->data = 3;
% G3 l0 T/ Q: M[color=rgba(0, 0, 0, 0.749019607843137)] n4->data = 4;
8 H# A7 f) a' X8 ^2 `[color=rgba(0, 0, 0, 0.749019607843137)] n5->data = 5;1 d2 C/ H, w |+ b' T# c7 t" W* n
[color=rgba(0, 0, 0, 0.749019607843137)] n6->data = 6;
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! x+ \7 a/ `, n% t[color=rgba(0, 0, 0, 0.749019607843137)] n1->left = n2; K$ [0 F3 }- n6 \$ F
[color=rgba(0, 0, 0, 0.749019607843137)] n1->right = n4;- m. y; _, d& t2 H2 [8 v; v
[color=rgba(0, 0, 0, 0.749019607843137)] n2->left = n3;
" \, Q0 E. w" p[color=rgba(0, 0, 0, 0.749019607843137)] n2->right = NULL;& I8 O7 e' d! }$ k* a/ b9 Y* @5 R
[color=rgba(0, 0, 0, 0.749019607843137)] n3->left = NULL;
3 `' Z5 m$ L5 L' X) j[color=rgba(0, 0, 0, 0.749019607843137)] n3->right = NULL;
6 y+ l7 @& D4 h' J5 s4 v[color=rgba(0, 0, 0, 0.749019607843137)] n4->left = n5;" N8 O+ v( ~8 J: N8 z r2 e7 _; d
[color=rgba(0, 0, 0, 0.749019607843137)] n4->right = n6;# m5 _, t/ A, ?1 [+ U
[color=rgba(0, 0, 0, 0.749019607843137)] n5->left = NULL;
8 N& t8 v* x8 e( B! Y; P5 O2 Z# R[color=rgba(0, 0, 0, 0.749019607843137)] n5->right = NULL;+ \: z/ D% Z% o% W/ ^
[color=rgba(0, 0, 0, 0.749019607843137)] n6->left = NULL;6 s1 G. s M, S% i
[color=rgba(0, 0, 0, 0.749019607843137)] n6->right = NULL;
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[color=rgba(0, 0, 0, 0.749019607843137)] return n1;, i) y0 r1 b4 @6 K+ I2 g
[color=rgba(0, 0, 0, 0.749019607843137)]}
% _: n# K+ z! r1 M7 R[color=rgba(0, 0, 0, 0.749019607843137)] s- W5 r6 `( N' Y" d3 x9 K
+ `& g7 D& P4 ], [8 g[color=rgba(0, 0, 0, 0.749019607843137)]int main()$ u* ~3 }( j) o+ ^8 K; x
[color=rgba(0, 0, 0, 0.749019607843137)]{% f0 U. a% A# a& @
[color=rgba(0, 0, 0, 0.749019607843137)] //先创建一个简单的二叉树结构* p8 p1 |1 }4 A$ g5 T: p
[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* root = CreateTree();% x8 i( g5 m1 L0 r$ w
[color=rgba(0, 0, 0, 0.749019607843137)]
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& u- v; v" j5 ^. w- S[color=rgba(0, 0, 0, 0.749019607843137)] //二叉树前序遍历
$ I, P9 k: z* w' |[color=rgba(0, 0, 0, 0.749019607843137)] printf("二叉树前序遍历:");4 o+ }. U1 s- D" N- Q7 Q
[color=rgba(0, 0, 0, 0.749019607843137)] PreOrder(root);! l0 V0 r1 Y6 I# U* K7 Q0 w% \9 b
[color=rgba(0, 0, 0, 0.749019607843137)] printf("\n");
: Z8 V& @6 \, a* B[color=rgba(0, 0, 0, 0.749019607843137)] //二叉树中序遍历3 C+ Y0 o! r: s( Z
[color=rgba(0, 0, 0, 0.749019607843137)] printf("二叉树中序序遍历:");
0 t' R) H. d- k[color=rgba(0, 0, 0, 0.749019607843137)] InOrder(root);
& b1 o+ x! k3 \[color=rgba(0, 0, 0, 0.749019607843137)] printf("\n");3 y7 k; C8 V6 R4 t3 Q" h1 m- {
[color=rgba(0, 0, 0, 0.749019607843137)] //二叉树后序遍历$ M) w, E; z8 E1 F; m8 O4 D9 Q
[color=rgba(0, 0, 0, 0.749019607843137)] printf("二叉树后序遍历:");' Z- d, |; A9 n" ?9 v4 y/ _- s1 g
[color=rgba(0, 0, 0, 0.749019607843137)] PostOrder(root);
* N: l$ ]+ m9 `4 G, k. q[color=rgba(0, 0, 0, 0.749019607843137)] printf("\n");8 T; Q! \% P" B M" }
[color=rgba(0, 0, 0, 0.749019607843137)]) u! E: ?" }7 C
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[color=rgba(0, 0, 0, 0.749019607843137)] return 0;# T( p9 W: L9 U2 n# o( t
[color=rgba(0, 0, 0, 0.749019607843137)]}
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[color=rgba(0, 0, 0, 0.749019607843137)]140 n" x8 m8 E+ c* K
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[color=rgba(0, 0, 0, 0.749019607843137)]17
1 {4 o% P8 {% }" g: A[color=rgba(0, 0, 0, 0.749019607843137)]18
5 k5 P- M* r! c. k3 ]/ r' a[color=rgba(0, 0, 0, 0.749019607843137)]19 T. j+ _* X7 @$ P
[color=rgba(0, 0, 0, 0.749019607843137)]20+ T9 ^6 g4 Y0 F5 q# J0 Z
[color=rgba(0, 0, 0, 0.749019607843137)]21
. D. v, n* E* o5 g[color=rgba(0, 0, 0, 0.749019607843137)]22 b, v6 S( ^; V. ~5 D( m' t
[color=rgba(0, 0, 0, 0.749019607843137)]231 ~) n* x+ U8 `9 I8 H
[color=rgba(0, 0, 0, 0.749019607843137)]24
m2 ?- K' M: }* t, H8 \4 v[color=rgba(0, 0, 0, 0.749019607843137)]25
$ W1 l3 C5 ~$ b3 a& `: I; t[color=rgba(0, 0, 0, 0.749019607843137)]26% b- l3 `1 @( F: M! D( G
[color=rgba(0, 0, 0, 0.749019607843137)]27( L3 f9 X3 j6 M* [7 g$ N
[color=rgba(0, 0, 0, 0.749019607843137)]28
# T* V# ]- b5 W$ i6 z5 _1 L[color=rgba(0, 0, 0, 0.749019607843137)]29
5 Q _- q$ y) m6 K$ H9 a# Z% V[color=rgba(0, 0, 0, 0.749019607843137)]30& [; ^7 W8 v4 n; O5 C5 C. H5 O2 F
[color=rgba(0, 0, 0, 0.749019607843137)]31
4 b" R" t$ Z9 k& g! J {[color=rgba(0, 0, 0, 0.749019607843137)]32
4 u- O+ J6 a( a% D: z[color=rgba(0, 0, 0, 0.749019607843137)]33( w8 Z# k3 d5 f4 Z3 ~% F7 f
[color=rgba(0, 0, 0, 0.749019607843137)]34
+ [% |7 F0 T9 D4 V[color=rgba(0, 0, 0, 0.749019607843137)]35" q, ]$ O: h( n% `9 Z
[color=rgba(0, 0, 0, 0.749019607843137)]36, y. x& B, R7 }0 I0 C) c9 d1 d" i
[color=rgba(0, 0, 0, 0.749019607843137)]37& O" h; k& F, w7 W/ {' Z
[color=rgba(0, 0, 0, 0.749019607843137)]38: S) W5 u' a% K) T8 z
[color=rgba(0, 0, 0, 0.749019607843137)]39& b6 f% Z- B+ k4 ?
[color=rgba(0, 0, 0, 0.749019607843137)]40% M3 p( T% j7 U( s& ]& a
[color=rgba(0, 0, 0, 0.749019607843137)]41
2 a2 f& `# l2 G[color=rgba(0, 0, 0, 0.749019607843137)]42, a* V$ E6 k9 b
[color=rgba(0, 0, 0, 0.749019607843137)]43
: y# a* z# C8 q* f) b[color=rgba(0, 0, 0, 0.749019607843137)]44
* X% J- ~8 Y6 j[color=rgba(0, 0, 0, 0.749019607843137)]45
$ U9 a h% M9 a( ?/ s[color=rgba(0, 0, 0, 0.749019607843137)]46
2 e1 w* u. f2 s4 Z# a$ R1 m[color=rgba(0, 0, 0, 0.749019607843137)]473 ?( S; r4 k) |9 l) L8 G
[color=rgba(0, 0, 0, 0.749019607843137)]48( ] Q- o) L2 q( U
[color=rgba(0, 0, 0, 0.749019607843137)]49
% H& T( o& w: l[color=rgba(0, 0, 0, 0.749019607843137)]50
+ g: [0 M# P! y# h4 `; M9 E* t[color=rgba(0, 0, 0, 0.749019607843137)]51. x) o$ U( z8 D
[color=rgba(0, 0, 0, 0.749019607843137)]52
, T- W: L. }9 @* |& X4 ~[color=rgba(0, 0, 0, 0.749019607843137)]531 T# x# v; q0 l" j/ O
[color=rgba(0, 0, 0, 0.749019607843137)]54
% m( ^& H; g8 [! C5 z[color=rgba(0, 0, 0, 0.749019607843137)]55, o+ U; d0 Y1 [9 O/ e/ [
[color=rgba(0, 0, 0, 0.749019607843137)]56
5 h. d- L; a4 |% D[color=rgba(0, 0, 0, 0.749019607843137)]57( J/ j) d3 m- `! P6 X/ x; f* r
[color=rgba(0, 0, 0, 0.749019607843137)]58( w( q, U7 ~) H6 w. I& o
[color=rgba(0, 0, 0, 0.749019607843137)]59
; V* _7 \- b: k: _. q Y[color=rgba(0, 0, 0, 0.749019607843137)]60: y' `$ X9 d+ z8 C7 z7 g
[color=rgba(0, 0, 0, 0.749019607843137)]61
( r8 P, H6 m/ F% z: c" P[color=rgba(0, 0, 0, 0.749019607843137)]62
* D g+ l! K+ J6 b/ S; g[color=rgba(0, 0, 0, 0.749019607843137)]63
6 }8 Z0 A0 @& R[color=rgba(0, 0, 0, 0.749019607843137)]646 I* n9 s/ D$ F {/ m8 U2 S
[color=rgba(0, 0, 0, 0.749019607843137)]657 o1 t c$ }3 s3 _% ?) O
[color=rgba(0, 0, 0, 0.749019607843137)]66
7 E* c/ p I. s+ W: A0 m[color=rgba(0, 0, 0, 0.749019607843137)]67
9 `# e u8 p. r3 l[color=rgba(0, 0, 0, 0.749019607843137)]68
' t$ M+ V( \. ?[color=rgba(0, 0, 0, 0.749019607843137)]69
' s; c1 u2 ^& V2 Y( q[color=rgba(0, 0, 0, 0.749019607843137)]70
" t3 J: M1 T- @7 D6 o$ c0 ~[color=rgba(0, 0, 0, 0.749019607843137)]71
7 z; G9 {9 `# u& R' j; Q/ d( Z[color=rgba(0, 0, 0, 0.749019607843137)]72
0 S. B/ b) v( C& |; h[color=rgba(0, 0, 0, 0.749019607843137)]73! U( U( A& G8 E9 s/ n7 X- O
[color=rgba(0, 0, 0, 0.749019607843137)]748 V( q ?- i; M( v
[color=rgba(0, 0, 0, 0.749019607843137)]75# U( }+ u3 q2 P$ f
[color=rgba(0, 0, 0, 0.749019607843137)]76
" K9 r' C0 ~+ `2 w6 u[color=rgba(0, 0, 0, 0.749019607843137)]774 _' E* R0 r$ V# j* Q8 j9 c' c+ ~
[color=rgba(0, 0, 0, 0.749019607843137)]784 m2 m n" t- [( m9 l7 {0 N" H3 G; ?
[color=rgba(0, 0, 0, 0.749019607843137)]79
/ |1 B/ p4 j9 R! D3 P[color=rgba(0, 0, 0, 0.749019607843137)]80
7 }/ W- G9 ]$ m! ]" \- _, n4 d[color=rgba(0, 0, 0, 0.749019607843137)]81
2 t6 ?5 i: [( M( q( u[color=rgba(0, 0, 0, 0.749019607843137)]82
4 b4 K! j, n) L$ a0 A[color=rgba(0, 0, 0, 0.749019607843137)]83
" b, d$ g# X6 z' m$ J, K0 S[color=rgba(0, 0, 0, 0.749019607843137)]84" J, u4 G7 C; g
[color=rgba(0, 0, 0, 0.749019607843137)]85
A$ m" e" S# L[color=rgba(0, 0, 0, 0.749019607843137)]86( T% f4 {( k U5 j$ d
[color=rgba(0, 0, 0, 0.749019607843137)]87# f' @" r/ E8 f9 x9 e; j. B
[color=rgba(0, 0, 0, 0.749019607843137)]88
+ ~& B& M {4 s[color=rgba(0, 0, 0, 0.749019607843137)]89/ R4 g' [; P& I- G2 W: ?
[color=rgba(0, 0, 0, 0.749019607843137)]906 Z. l$ p7 f/ [
[color=rgba(0, 0, 0, 0.749019607843137)]91 F# k! j0 S9 j9 x
[color=rgba(0, 0, 0, 0.749019607843137)]92" W7 R2 k9 p! N
[color=rgba(0, 0, 0, 0.749019607843137)]939 j% ~ D2 h3 ?# G+ D! x
[color=rgba(0, 0, 0, 0.749019607843137)]94 k- V+ Q# d4 n
[color=rgba(0, 0, 0, 0.749019607843137)]95
& A. k' ?+ J }# C! l) C R( w: o[color=rgba(0, 0, 0, 0.749019607843137)]96( e$ |0 A3 r8 i4 M
[color=rgba(0, 0, 0, 0.749019607843137)]97
8 y: Z+ X" [ P# S8 S' u[color=rgba(0, 0, 0, 0.749019607843137)]98
: `! W! l" z! c! d) I[color=rgba(0, 0, 0, 0.749019607843137)]99
2 ^, l* w$ Y' r4 A/ Z9 D3 P5 [[color=rgba(0, 0, 0, 0.749019607843137)]100$ u& W' A: W, f( {
[color=rgba(0, 0, 0, 0.749019607843137)]101
6 s6 x8 a+ E0 A$ _$ Q[color=rgba(0, 0, 0, 0.749019607843137)]102
2 s" e8 }) s _ b" g- ~[color=rgba(0, 0, 0, 0.749019607843137)]1037 R5 S& ?7 O j$ U- \3 y( G
[color=rgba(0, 0, 0, 0.749019607843137)]104; o) k O2 A2 j3 G0 a
[color=rgba(0, 0, 0, 0.749019607843137)]1053 S% w3 e: [9 h1 I, @
[color=rgba(0, 0, 0, 0.749019607843137)]106
6 q, R9 o1 A' U7 d# g3 M[color=rgba(0, 0, 0, 0.749019607843137)]107
: V6 i* ?+ L6 }1 a[color=rgba(0, 0, 0, 0.749019607843137)]108
; ^, X. H8 K+ i: Y& b) n[color=rgba(0, 0, 0, 0.749019607843137)]1090 n8 \/ Z2 g- m( X1 Y
[color=rgba(0, 0, 0, 0.749019607843137)]110
: K y1 f* C. T1 A! }, a! x0 M[color=rgba(0, 0, 0, 0.749019607843137)]111
+ Q5 B1 X* S9 L- k- t* q[color=rgba(0, 0, 0, 0.749019607843137)]测试结果：. }- P) F$ K3 ~% V+ F
[color=rgba(0, 0, 0, 0.749019607843137)]
# b" O1 q+ I: N8 w1 U- C) D/ u" K3 k+ E! I
[color=rgba(0, 0, 0, 0.749019607843137)]7 U/ x' J/ n3 _6 b2 B1 Y8 y
4 X5 o) @4 w1 J3 d, i. c
[color=rgba(0, 0, 0, 0.749019607843137)]2.2计算二叉树的大小9 J/ E- n( H$ k* X
[color=rgba(0, 0, 0, 0.749019607843137)]时间复杂度为O(N)
5 B# h; Y& I/ o1 B[color=rgba(0, 0, 0, 0.749019607843137)]' a* z# A7 M$ J" Z* S
: Z7 X5 t& \& I" M5 i& O[color=rgba(0, 0, 0, 0.749019607843137)]//二叉树的大小
$ E8 w4 T' R# z8 Q. C0 {[color=rgba(0, 0, 0, 0.749019607843137)]//法一：全局变量
8 W" A* k# I" l$ m1 H8 }7 e- s1 ?[color=rgba(0, 0, 0, 0.749019607843137)]//int count = 0;6 G3 K: @% _4 |, b F
[color=rgba(0, 0, 0, 0.749019607843137)]//void TreeSize(BTNode* root)
: y2 h1 Q' b4 W$ W[color=rgba(0, 0, 0, 0.749019607843137)]//{2 u/ v4 X- }+ m0 J
[color=rgba(0, 0, 0, 0.749019607843137)]// if (root == NULL)
. K8 S: x# h& H- } Y6 h! `8 L[color=rgba(0, 0, 0, 0.749019607843137)]// {
7 h9 r5 O* R5 x( q3 p[color=rgba(0, 0, 0, 0.749019607843137)]// return;
' F% W6 t; w" ?" F[color=rgba(0, 0, 0, 0.749019607843137)]// }
/ y4 u, t B% B6 u' g[color=rgba(0, 0, 0, 0.749019607843137)]// count++;
$ D. E: m$ P' d( v$ H: l4 N[color=rgba(0, 0, 0, 0.749019607843137)]// TreeSize(root->left);& B) ]# h8 h* Q
[color=rgba(0, 0, 0, 0.749019607843137)]// TreeSize(root->right);
' H3 s/ G! q# d3 n0 G+ p: L[color=rgba(0, 0, 0, 0.749019607843137)]//9 c; y+ j. i z3 D: b
[color=rgba(0, 0, 0, 0.749019607843137)]// return;//函数栈帧层层返回，最后回到根节点1，结束了1的右子树函数，什么都不返回，因为count是全局变量
|- v; U" W: D+ k# h[color=rgba(0, 0, 0, 0.749019607843137)]//}# V) R0 B% V) {1 |/ w) j
[color=rgba(0, 0, 0, 0.749019607843137)]//法二：子问题思路：分而治之* F+ }, H" @# g
[color=rgba(0, 0, 0, 0.749019607843137)]int TreeSize(BTNode* root); c9 S0 H+ J7 Y% g0 C5 J2 b% a
[color=rgba(0, 0, 0, 0.749019607843137)]{9 U% U: |! a% O
[color=rgba(0, 0, 0, 0.749019607843137)] return root == NULL ? 0 : TreeSize(root->left) + TreeSize(root->right) + 1;
) Q- n. v" C; b$ `2 f[color=rgba(0, 0, 0, 0.749019607843137)]}
7 B; J9 a6 j7 Y0 @% C[color=rgba(0, 0, 0, 0.749019607843137)]1
0 [3 B5 V7 } o2 r$ W, [[color=rgba(0, 0, 0, 0.749019607843137)]2$ Q9 o8 Z! V" \
[color=rgba(0, 0, 0, 0.749019607843137)]3
' a7 u( O( q9 c) Q[color=rgba(0, 0, 0, 0.749019607843137)]4. }- P7 m9 I( O$ T# g2 {
[color=rgba(0, 0, 0, 0.749019607843137)]51 Q2 V! S5 Z7 G
[color=rgba(0, 0, 0, 0.749019607843137)]6" z0 l/ @0 H) X: a V1 ^
[color=rgba(0, 0, 0, 0.749019607843137)]7$ r( h; |$ V$ ]/ o
[color=rgba(0, 0, 0, 0.749019607843137)]8
/ z" m" Y$ M7 B9 T% f. m[color=rgba(0, 0, 0, 0.749019607843137)]9% j; U& h. \+ D
[color=rgba(0, 0, 0, 0.749019607843137)]10
% T- S8 m9 V% e[color=rgba(0, 0, 0, 0.749019607843137)]11/ K; @; u. N# j6 _7 |, J
[color=rgba(0, 0, 0, 0.749019607843137)]12
' D' Q# u! @& d5 h# U[color=rgba(0, 0, 0, 0.749019607843137)]13( X* v; ]0 i: B2 ]$ T2 {
[color=rgba(0, 0, 0, 0.749019607843137)]14
1 n# A% E/ g' m/ r0 T5 c, c[color=rgba(0, 0, 0, 0.749019607843137)]15
+ G' T1 v) ?+ i/ i4 ? X) `[color=rgba(0, 0, 0, 0.749019607843137)]16
0 X9 A1 j- B0 {# Q9 s* ?[color=rgba(0, 0, 0, 0.749019607843137)]17
3 C4 C8 v% v7 B8 E1 p* g[color=rgba(0, 0, 0, 0.749019607843137)]18
- R6 u" i& i/ i[color=rgba(0, 0, 0, 0.749019607843137)]19/ D3 P% H- F/ U
[color=rgba(0, 0, 0, 0.749019607843137)]20
! H! m+ N: u! x9 U3 w[color=rgba(0, 0, 0, 0.749019607843137)]2.3计算二叉树叶子结点的个数4 [! L& _9 C) b3 \" y4 l
[color=rgba(0, 0, 0, 0.749019607843137)]//计算二叉树叶子结点的个数
6 z( B! p* U' \! e[color=rgba(0, 0, 0, 0.749019607843137)]int TreeLeafSize(BTNode* root)5 O% Q% p; ~/ c& I( O( ?
[color=rgba(0, 0, 0, 0.749019607843137)]{: U, h6 q( Q8 {* a( T
[color=rgba(0, 0, 0, 0.749019607843137)] if (root == NULL)//首先得考虑空树的情况,0个叶子结点! s' P8 _5 S& X/ @7 u. H. w
[color=rgba(0, 0, 0, 0.749019607843137)] {$ T1 a5 _, r% K" [
[color=rgba(0, 0, 0, 0.749019607843137)] return 0;( M& i4 L! F- Q9 [
[color=rgba(0, 0, 0, 0.749019607843137)] }3 ~+ g) V* d* _6 E; z
[color=rgba(0, 0, 0, 0.749019607843137)] //叶子结点的特征就是左右子树为空
# C. D8 x# h8 ~" W0 J[color=rgba(0, 0, 0, 0.749019607843137)] if (root->left == NULL && root->right == NULL), [1 b1 v) s# G
[color=rgba(0, 0, 0, 0.749019607843137)] {( c. i+ y8 @% b8 C- X# ?
[color=rgba(0, 0, 0, 0.749019607843137)] return 1;
- S+ ^# i* ^7 L2 @[color=rgba(0, 0, 0, 0.749019607843137)] }4 Z- V. W" W% k5 ^, h
[color=rgba(0, 0, 0, 0.749019607843137)] return TreeLeafSize(root->left) + TreeLeafSize(root->right);# a& C4 a0 B4 T9 |! E
[color=rgba(0, 0, 0, 0.749019607843137)]}
- l2 a" u T7 X[color=rgba(0, 0, 0, 0.749019607843137)]1
B" T5 I+ Y2 s[color=rgba(0, 0, 0, 0.749019607843137)]2
* q$ h0 J7 ]6 ?; T# W& @[color=rgba(0, 0, 0, 0.749019607843137)]3
' i3 ^* R8 M) U2 I[color=rgba(0, 0, 0, 0.749019607843137)]44 _8 G; \0 v4 O) n
[color=rgba(0, 0, 0, 0.749019607843137)]5. w, Z+ G* e" Q% @6 ? m
[color=rgba(0, 0, 0, 0.749019607843137)]6; F5 H1 K8 c1 s9 ~0 R: u; \; k$ @
[color=rgba(0, 0, 0, 0.749019607843137)]72 E4 y, |) W- A6 n6 j
[color=rgba(0, 0, 0, 0.749019607843137)]8
4 i4 [+ d- J3 C8 R5 X[color=rgba(0, 0, 0, 0.749019607843137)]9: S, y/ P) h) J$ n+ k& z
[color=rgba(0, 0, 0, 0.749019607843137)]10
# n7 k# D' M! i/ m. T2 l4 U, z! s[color=rgba(0, 0, 0, 0.749019607843137)]11
. u; m% M/ X1 X( k4 \[color=rgba(0, 0, 0, 0.749019607843137)]12
, z1 X, \7 p2 o7 F9 @# V[color=rgba(0, 0, 0, 0.749019607843137)]13
" o" r9 |( h1 g) F$ m[color=rgba(0, 0, 0, 0.749019607843137)]14
( g- P+ F( r& c0 _. f* t5 H) j[color=rgba(0, 0, 0, 0.749019607843137)]2.4计算二叉树的高度5 ?4 \4 C) K1 p; a8 m( ~+ H+ \
[color=rgba(0, 0, 0, 0.749019607843137)]int TreeHeight(BTNode* root)
: D% r7 P# d( M" Y; K% w[color=rgba(0, 0, 0, 0.749019607843137)]{
% q: g. ]9 l% j+ v8 ?[color=rgba(0, 0, 0, 0.749019607843137)] //空树高度为0( w8 u7 [8 F0 W k
[color=rgba(0, 0, 0, 0.749019607843137)] if (root == NULL)
3 v& K5 K& z1 J+ M5 L7 A4 Q[color=rgba(0, 0, 0, 0.749019607843137)] {
" O, L D( x$ G; X6 g[color=rgba(0, 0, 0, 0.749019607843137)] return 0;
! `6 f6 f& p9 ~ x& C2 f[color=rgba(0, 0, 0, 0.749019607843137)] }/ D! v# a/ ^/ z* u2 D9 ^3 z
[color=rgba(0, 0, 0, 0.749019607843137)] //树的高度是较高的那棵子树
9 @! Z# p. {+ j/ D) S' [[color=rgba(0, 0, 0, 0.749019607843137)] int lh = TreeHeight(root->left);//左子树的高度
1 O" b) R1 L2 Q. O/ V" s. f3 p& J[color=rgba(0, 0, 0, 0.749019607843137)] int rh = TreeHeight(root->right);//右子树的高度
+ s; s- {/ @2 k5 E8 ~* y% J3 l[color=rgba(0, 0, 0, 0.749019607843137)]( j5 R8 Q) Y3 k0 d2 v3 ~
, B \* @! }, r[color=rgba(0, 0, 0, 0.749019607843137)] return lh > rh ? lh + 1 : rh + 1;/ q+ w% b3 |4 F
[color=rgba(0, 0, 0, 0.749019607843137)]}6 Y2 ~+ H. G7 Z; e; R4 j
[color=rgba(0, 0, 0, 0.749019607843137)]1
3 |4 N0 H4 D+ c[color=rgba(0, 0, 0, 0.749019607843137)]2
+ E p$ D" y2 {7 N, c[color=rgba(0, 0, 0, 0.749019607843137)]3% J8 d3 a5 y+ I
[color=rgba(0, 0, 0, 0.749019607843137)]4
# c% q2 y' ]6 ~: r- n0 Z1 B[color=rgba(0, 0, 0, 0.749019607843137)]5
, K$ x8 n1 | h[color=rgba(0, 0, 0, 0.749019607843137)]6" W- L1 S" P( A& b8 R
[color=rgba(0, 0, 0, 0.749019607843137)]77 t3 k2 ^0 }/ i S+ X' N0 q/ Y
[color=rgba(0, 0, 0, 0.749019607843137)]8+ |9 ~, Z8 S. }, `% v
[color=rgba(0, 0, 0, 0.749019607843137)]9' k, s. E8 ]) z6 x
[color=rgba(0, 0, 0, 0.749019607843137)]10
) y* k# T! x, t F[color=rgba(0, 0, 0, 0.749019607843137)]113 l" Y. R& H; ~" ~8 u
[color=rgba(0, 0, 0, 0.749019607843137)]12
; b F P3 j. D8 M( |; q' T% y4 s[color=rgba(0, 0, 0, 0.749019607843137)]135 x2 l. b* ]7 E0 m# J
[color=rgba(0, 0, 0, 0.749019607843137)]2.5计算第K层结点的个数
4 C4 h/ u% F' C/ i6 N5 S1 A' u[color=rgba(0, 0, 0, 0.749019607843137)]" z) M7 W" a5 `4 l0 @# o- L% z
( h& \8 L2 T1 L, w[color=rgba(0, 0, 0, 0.749019607843137)]
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* s8 l. N* J0 k: r3 s[color=rgba(0, 0, 0, 0.749019607843137)]求第K层的结点个数，转换成求子树第K-1层的结点个数。举个栗子：如果我们要求这棵二叉树第3层的结点个数（为3），就转换成求左子树根结点2的第2层的结点个数+右子树4第二层的结点个数。。。3 V5 y- k2 F# b" e2 q
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[color=rgba(0, 0, 0, 0.749019607843137)]//计算第K层结点的个数
; x2 H5 z" |3 }! X; z[color=rgba(0, 0, 0, 0.749019607843137)]int TreeLevel(BTNode* root,int K)
2 [: {. u4 j$ s/ F, u) G5 C[color=rgba(0, 0, 0, 0.749019607843137)]{3 c8 s2 ~! Z+ k$ l2 [- X
[color=rgba(0, 0, 0, 0.749019607843137)] assert(K > 0);: Q/ M, j5 {# _0 i$ o. l
[color=rgba(0, 0, 0, 0.749019607843137)] if (root == NULL)$ j; P. R# S: }7 _
[color=rgba(0, 0, 0, 0.749019607843137)] {
% W6 e1 |0 J6 c: w* C8 A d[color=rgba(0, 0, 0, 0.749019607843137)] return 0;
2 u* T( F" b$ `$ Y. P9 B[color=rgba(0, 0, 0, 0.749019607843137)] }" h: H8 q/ w+ U; u+ B- I* l
[color=rgba(0, 0, 0, 0.749019607843137)] //如果是第一层(递归出口). Q- K) g% l) ]
[color=rgba(0, 0, 0, 0.749019607843137)] if (K == 1)
# q' }) O. _3 D2 _# k$ M$ X[color=rgba(0, 0, 0, 0.749019607843137)] {
) s; Q3 {* x- a$ e: p2 Z[color=rgba(0, 0, 0, 0.749019607843137)] return 1;
+ M2 [3 Z* [$ i5 W% \[color=rgba(0, 0, 0, 0.749019607843137)] }4 }( ~5 ` H7 w3 n) N
[color=rgba(0, 0, 0, 0.749019607843137)] //转换成子树的第K-1层
9 Z/ |( D( a F! k& n' K[color=rgba(0, 0, 0, 0.749019607843137)] return TreeLevel(root->left,K-1) + TreeLevel(root->right,K-1);/ C8 i9 B3 X: Y1 T' d% U. a( @
[color=rgba(0, 0, 0, 0.749019607843137)]}
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[color=rgba(0, 0, 0, 0.749019607843137)]3
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[color=rgba(0, 0, 0, 0.749019607843137)]6
/ o9 ]6 F v+ k0 P6 K[color=rgba(0, 0, 0, 0.749019607843137)]7
5 f* B- u3 v3 Y: l9 |5 J" @( U8 w4 B& }! j[color=rgba(0, 0, 0, 0.749019607843137)]89 q# J5 M% E0 J7 ]
[color=rgba(0, 0, 0, 0.749019607843137)]95 @+ U& X1 M4 \9 k/ D3 R2 U+ |/ ^
[color=rgba(0, 0, 0, 0.749019607843137)]10
3 A+ S' O1 |( f" [" O5 f A[color=rgba(0, 0, 0, 0.749019607843137)]114 K. d4 `0 j, b; u/ j$ l' O8 N
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[color=rgba(0, 0, 0, 0.749019607843137)]14
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[color=rgba(0, 0, 0, 0.749019607843137)]161 B- v% r: Q8 P0 Y( T1 d1 |1 ?6 x
[color=rgba(0, 0, 0, 0.749019607843137)]2.6二叉树查找3 Q( a' A, g6 ^9 B" g
[color=rgba(0, 0, 0, 0.749019607843137)]//二叉树查找, |+ Z; D$ z4 J8 n$ q' O
[color=rgba(0, 0, 0, 0.749019607843137)]BTNode* TreeFind(BTNode* root, BTDataType data)# I7 y) Y, h+ M2 w+ ?; x
[color=rgba(0, 0, 0, 0.749019607843137)]{
) b( A5 h2 L$ O4 b[color=rgba(0, 0, 0, 0.749019607843137)] if (root == NULL)
# i0 b$ ?. P3 y) {3 w, P7 E[color=rgba(0, 0, 0, 0.749019607843137)] {
, Z! R, Z& T% r[color=rgba(0, 0, 0, 0.749019607843137)] return NULL;) N" i1 t* A2 y" E7 {
[color=rgba(0, 0, 0, 0.749019607843137)] }
8 k2 A" ?: ]( F: V$ c# D3 N8 z[color=rgba(0, 0, 0, 0.749019607843137)] if (root->data == data)
+ T( d0 V6 [+ C: }+ x) J C1 o1 t[color=rgba(0, 0, 0, 0.749019607843137)] {" |* F4 }9 j2 M$ y# W
[color=rgba(0, 0, 0, 0.749019607843137)] return root;
6 j4 t- }2 u% c9 {; F0 [[color=rgba(0, 0, 0, 0.749019607843137)] }. d: a$ Q9 `3 [8 B0 }3 S
[color=rgba(0, 0, 0, 0.749019607843137)] //先查找左子树
/ G5 _- E1 j8 X( Q( l: T& h[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* lret = TreeFind(root->left, data);
/ [1 X- ~% ^! q9 |. ^+ a[color=rgba(0, 0, 0, 0.749019607843137)] if (lret)
, `6 a5 _& x; w3 d6 b9 a! M[color=rgba(0, 0, 0, 0.749019607843137)] return lret;
/ K" n Y' P* o1 N& I9 [# ]! b1 o- `; ~8 Q: B[color=rgba(0, 0, 0, 0.749019607843137)] //再查找右子树8 E+ s# `$ n* p& ]1 q/ P, M6 D
[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* rret = TreeFind(root->right, data);/ y+ J) |3 ~8 g2 R2 S
[color=rgba(0, 0, 0, 0.749019607843137)] if (rret)
/ C! g H8 v7 R[color=rgba(0, 0, 0, 0.749019607843137)] return rret;# |2 z8 L' ]8 K$ i
[color=rgba(0, 0, 0, 0.749019607843137)] return NULL;' M% c: h3 P3 |% x$ Q
[color=rgba(0, 0, 0, 0.749019607843137)]}& X+ V7 ?9 [3 e2 `$ L; o
[color=rgba(0, 0, 0, 0.749019607843137)]————————————————
+ S3 F$ j- i" t M, f1 V[color=rgba(0, 0, 0, 0.749019607843137)]版权声明：本文为CSDN博主「SouLinya」的原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接及本声明。
! V3 }9 S3 T/ V[color=rgba(0, 0, 0, 0.749019607843137)]原文链接：https://blog.csdn.net/weixin_63449996/article/details/126841212. h& P$ k) {) v3 n) k7 R
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