【数据结构】二叉树的遍历：前序，中序，后序的递归结构遍历
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4 A% Q$ A" \ x! F M0 ~[color=rgba(0, 0, 0, 0.749019607843137)]文章目录
2 r2 Y1 m, l3 z* m$ \# J[color=rgba(0, 0, 0, 0.749019607843137)]前言
; W% e- n: B2 l: F3 d[color=rgba(0, 0, 0, 0.749019607843137)]1.二叉树的遍历方式
6 a0 r h9 |" Z4 S* w[color=rgba(0, 0, 0, 0.749019607843137)]2.二叉树的遍历及相关函数（代码实现）, i5 h' ]6 B* E8 [& m
[color=rgba(0, 0, 0, 0.749019607843137)]2.1前序/中序/后序的遍历
4 B* y: E# q0 M; }& M$ t5 D[color=rgba(0, 0, 0, 0.749019607843137)]2.2计算二叉树的大小- g, \2 E$ d. c# w
[color=rgba(0, 0, 0, 0.749019607843137)]2.3计算二叉树叶子结点的个数4 Y$ U$ i7 ~7 e& t0 Z/ o
[color=rgba(0, 0, 0, 0.749019607843137)]2.4计算二叉树的高度
9 U" |5 S7 [+ e[color=rgba(0, 0, 0, 0.749019607843137)]2.5计算第K层结点的个数
8 x: [" k* A2 H& J[color=rgba(0, 0, 0, 0.749019607843137)]2.6二叉树查找
, X: }; v: D& P1 `2 M: F[color=rgba(0, 0, 0, 0.749019607843137)]前言
2 G$ a' ?& X! O4 K4 \9 W% P[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。( _0 f1 V4 W# o2 k$ ~% N
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f! q" F' U1 l/ ?0 U[color=rgba(0, 0, 0, 0.749019607843137)]1.二叉树的遍历方式
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7 g ?- z9 N' _' w% g" u1 K' V[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： {; K" Q2 @, |. J- [9 L
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9 c: }7 e/ z( |7 X6 n, Z: E5 Y[color=rgba(0, 0, 0, 0.749019607843137)]1. 前序遍历（先序，先根):根——左子树——右子树 w& D- n& ]& ?& X' U; A) C
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) g4 b) @& s5 `[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
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; Y n/ f0 M5 Y! _[color=rgba(0, 0, 0, 0.749019607843137)]2.二叉树的遍历及相关函数（代码实现） \& {, `6 U& N ~4 d9 ^
[color=rgba(0, 0, 0, 0.749019607843137)]思路：分而治之
" y% e# a8 C! V" @[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；
" c8 D- V/ [7 a) l8 b4 S[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；* h$ C6 ?, | L# _4 f1 g/ A
[color=rgba(0, 0, 0, 0.749019607843137)]3.尝试计算这个二叉树的大小（利用递归）；
2 N5 e# Z! R% u& m9 {' R' `- ?: e[color=rgba(0, 0, 0, 0.749019607843137)]4.尝试计算叶子结点的个数（利用递归）；- J( f9 ?) {% e1 @" D8 E: F
[color=rgba(0, 0, 0, 0.749019607843137)]5.尝试计算二叉树的高度（利用递归）；
, j2 m' d( _$ P$ A1 L" H( q[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）；8 O6 k6 J7 @& M' ]+ C
[color=rgba(0, 0, 0, 0.749019607843137)]7.尝试写出二叉树查找的函数（利用递归）。
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[color=rgba(0, 0, 0, 0.749019607843137)]2.1前序/中序/后序的遍历8 a# D) V! r6 V; e/ B4 h
[color=rgba(0, 0, 0, 0.749019607843137)]#define _CRT_SECURE_NO_WARNINGS 1
6 j) x0 z* e" @) r9 u3 l[color=rgba(0, 0, 0, 0.749019607843137)]#include<stdio.h>
- M# z% B0 E: t7 V[color=rgba(0, 0, 0, 0.749019607843137)]#include<assert.h>
3 {6 Z: O9 J5 a' G. B[color=rgba(0, 0, 0, 0.749019607843137)]#include<stdlib.h>
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[color=rgba(0, 0, 0, 0.749019607843137)]typedef int BTDataType;
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[color=rgba(0, 0, 0, 0.749019607843137)]//定义二叉树结点的结构体
- J- Z& b9 n& L6 s8 Z[color=rgba(0, 0, 0, 0.749019607843137)]typedef struct BinaryTreeNode5 l+ ?5 S8 f: J; ?0 Z; T p
[color=rgba(0, 0, 0, 0.749019607843137)]{: q- w8 ?! B! r# w4 C M' O
[color=rgba(0, 0, 0, 0.749019607843137)] BTDataType data; r$ f H, l1 U* D
[color=rgba(0, 0, 0, 0.749019607843137)] struct BinaryTreeNode* left;
& y( h- f9 Y; [; b8 \# A[color=rgba(0, 0, 0, 0.749019607843137)] struct BinaryTreeNode* right;( t8 i% G+ R) |: s. a: B# F
[color=rgba(0, 0, 0, 0.749019607843137)]}BTNode;/ E1 G0 U. z4 w1 |7 A
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: B4 Z5 a2 i, w) x) H[color=rgba(0, 0, 0, 0.749019607843137)]//前序遍历* L0 c1 T' I- i
[color=rgba(0, 0, 0, 0.749019607843137)]void PreOrder(BTNode* root)
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[color=rgba(0, 0, 0, 0.749019607843137)] if (root == NULL)! k1 m1 T; b8 [& u
[color=rgba(0, 0, 0, 0.749019607843137)] {
& C+ |/ R) X9 c& d ][color=rgba(0, 0, 0, 0.749019607843137)] printf("NULL ");
0 k9 g# N' G* I* I[color=rgba(0, 0, 0, 0.749019607843137)] return;$ Y% R" ]4 Y) a, q- Y3 j: `
[color=rgba(0, 0, 0, 0.749019607843137)] }
. l2 C( P8 F# V# R% j m[color=rgba(0, 0, 0, 0.749019607843137)] printf("%d ", root->data); L# i5 Y* ^$ \: I& t1 l+ @
[color=rgba(0, 0, 0, 0.749019607843137)] PreOrder(root->left);2 _$ O% i3 r) | n, _
[color=rgba(0, 0, 0, 0.749019607843137)] PreOrder(root->right);
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, V3 w3 L) Y9 M q1 @[color=rgba(0, 0, 0, 0.749019607843137)]//中序遍历, W2 a: r; y6 p/ A4 a( Z
[color=rgba(0, 0, 0, 0.749019607843137)]void InOrder(BTNode* root) ~% d8 X; u& x% e: T4 y0 n
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[color=rgba(0, 0, 0, 0.749019607843137)] if (root == NULL)
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[color=rgba(0, 0, 0, 0.749019607843137)] printf("NULL ");# S; O. s* `# o. ~
[color=rgba(0, 0, 0, 0.749019607843137)] return;
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# d; r* x+ b4 B! @8 I$ J' q0 z( S[color=rgba(0, 0, 0, 0.749019607843137)] InOrder(root->left);5 |3 }) D0 @0 g# U& J9 D. c
[color=rgba(0, 0, 0, 0.749019607843137)] printf("%d ", root->data);
+ p" ?/ i/ N3 [; P: U[color=rgba(0, 0, 0, 0.749019607843137)] InOrder(root->right);
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9 K. ~7 [3 I1 {) s$ [[color=rgba(0, 0, 0, 0.749019607843137)]//后序遍历
# o' m/ U" |9 b% E# a[color=rgba(0, 0, 0, 0.749019607843137)]void PostOrder(BTNode* root)
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[color=rgba(0, 0, 0, 0.749019607843137)] if (root == NULL)' W# `1 R+ }. w
[color=rgba(0, 0, 0, 0.749019607843137)] {' X* v* K$ x- q) _: r4 W
[color=rgba(0, 0, 0, 0.749019607843137)] printf("NULL ");5 g( b8 Y/ }7 i
[color=rgba(0, 0, 0, 0.749019607843137)] return;
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[color=rgba(0, 0, 0, 0.749019607843137)] PostOrder(root->left);7 F$ Q, `: \4 @6 w/ t# p3 G
[color=rgba(0, 0, 0, 0.749019607843137)] PostOrder(root->right);
: s- _$ Y$ {0 C2 U[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
+ ?- B x4 o3 i[color=rgba(0, 0, 0, 0.749019607843137)]BTNode* CreateTree()4 K7 ~- {; u$ }/ J D6 @
[color=rgba(0, 0, 0, 0.749019607843137)]{' o4 N8 M6 O: _* x1 d' Y/ L+ P! V
[color=rgba(0, 0, 0, 0.749019607843137)] //先动态开辟6个结点的空间
4 S6 c8 Y, M; Z3 Y[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* n1 = (BTNode*)malloc(sizeof(BTNode));
7 M) ?, ?4 N$ H. s% U[color=rgba(0, 0, 0, 0.749019607843137)] assert(n1);
: q* r$ B) j* l" w) ~' \[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* n2 = (BTNode*)malloc(sizeof(BTNode));4 W" n7 F# J2 i$ r
[color=rgba(0, 0, 0, 0.749019607843137)] assert(n2);: V8 q) ?2 X1 `6 r7 B: n6 Q
[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* n3 = (BTNode*)malloc(sizeof(BTNode));
5 X7 G/ @# O. w5 l- C+ r( h[color=rgba(0, 0, 0, 0.749019607843137)] assert(n3);$ D+ z6 d$ V. ?* l
[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* n4 = (BTNode*)malloc(sizeof(BTNode));4 E7 j# T. U- f$ G( P5 |; b
[color=rgba(0, 0, 0, 0.749019607843137)] assert(n4);- l0 R3 v" O* \' g9 r% l; x# |* A7 ^
[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* n5 = (BTNode*)malloc(sizeof(BTNode));& }6 ?. U; \; O! y4 S
[color=rgba(0, 0, 0, 0.749019607843137)] assert(n5);5 C& G* t4 _7 Z! r) @+ Q6 X
[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* n6 = (BTNode*)malloc(sizeof(BTNode));
. P" o1 C' s8 P+ a+ P- ^[color=rgba(0, 0, 0, 0.749019607843137)] assert(n6);
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[color=rgba(0, 0, 0, 0.749019607843137)] n1->data = 1;% ^5 N# C! M- g) j! H
[color=rgba(0, 0, 0, 0.749019607843137)] n2->data = 2;2 W+ |6 C! v# Z
[color=rgba(0, 0, 0, 0.749019607843137)] n3->data = 3;6 `3 M! k1 V, O
[color=rgba(0, 0, 0, 0.749019607843137)] n4->data = 4;
8 K) ^. [# e- U/ L' ^[color=rgba(0, 0, 0, 0.749019607843137)] n5->data = 5;9 t) q* \, C5 P! @
[color=rgba(0, 0, 0, 0.749019607843137)] n6->data = 6;( g4 x/ B3 G1 A. E3 ^
[color=rgba(0, 0, 0, 0.749019607843137)]
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[color=rgba(0, 0, 0, 0.749019607843137)] n1->left = n2;
& ]" U, f; R# c S l$ Y% p- G5 Y[color=rgba(0, 0, 0, 0.749019607843137)] n1->right = n4;- X. l% B0 w" N. J1 j# m
[color=rgba(0, 0, 0, 0.749019607843137)] n2->left = n3;# j1 e* r+ u% I/ R
[color=rgba(0, 0, 0, 0.749019607843137)] n2->right = NULL;7 D- \1 p: ~% R1 c) U5 }
[color=rgba(0, 0, 0, 0.749019607843137)] n3->left = NULL;
/ ~9 w c# Y1 ?6 c! @6 `[color=rgba(0, 0, 0, 0.749019607843137)] n3->right = NULL;
5 W# g% ^8 P5 b: q; K[color=rgba(0, 0, 0, 0.749019607843137)] n4->left = n5;0 v8 q7 ?5 a6 H( _6 A% q
[color=rgba(0, 0, 0, 0.749019607843137)] n4->right = n6;& ^3 C% j7 g' {6 a& j
[color=rgba(0, 0, 0, 0.749019607843137)] n5->left = NULL;
; d3 Z% E& m* |4 X[color=rgba(0, 0, 0, 0.749019607843137)] n5->right = NULL;1 ]; r) J4 j* Z: }- h, J
[color=rgba(0, 0, 0, 0.749019607843137)] n6->left = NULL;
6 k) ]9 d' u% ^! E! Z- v& m[color=rgba(0, 0, 0, 0.749019607843137)] n6->right = NULL;8 s; W) _; m2 w; i
[color=rgba(0, 0, 0, 0.749019607843137)]
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[color=rgba(0, 0, 0, 0.749019607843137)] return n1;6 f* P/ f$ H R
[color=rgba(0, 0, 0, 0.749019607843137)]}
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# O+ o2 i3 U0 E/ l a3 `[color=rgba(0, 0, 0, 0.749019607843137)]int main(). m+ f. s+ G1 a; a" Z
[color=rgba(0, 0, 0, 0.749019607843137)]{( h( m& l; Y* h/ U
[color=rgba(0, 0, 0, 0.749019607843137)] //先创建一个简单的二叉树结构
- ^# u% l; W8 }& o) W0 j[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* root = CreateTree();
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[color=rgba(0, 0, 0, 0.749019607843137)] //二叉树前序遍历( |1 T, w" T [% C; `
[color=rgba(0, 0, 0, 0.749019607843137)] printf("二叉树前序遍历:");8 C; @7 {! l8 e6 Z
[color=rgba(0, 0, 0, 0.749019607843137)] PreOrder(root);
* ?# C/ P/ Q, z6 G& J[color=rgba(0, 0, 0, 0.749019607843137)] printf("\n");# C$ x3 g; b5 T# i9 ~
[color=rgba(0, 0, 0, 0.749019607843137)] //二叉树中序遍历
3 l; D( N6 A$ }/ |( D[color=rgba(0, 0, 0, 0.749019607843137)] printf("二叉树中序序遍历:");% _0 o) A3 ?5 z$ y+ m7 ~( V
[color=rgba(0, 0, 0, 0.749019607843137)] InOrder(root);1 e2 v6 o3 a7 ~* X
[color=rgba(0, 0, 0, 0.749019607843137)] printf("\n");& m8 L: D+ l. b4 }3 s
[color=rgba(0, 0, 0, 0.749019607843137)] //二叉树后序遍历
3 c5 n2 [" b7 x% K! b9 Q[color=rgba(0, 0, 0, 0.749019607843137)] printf("二叉树后序遍历:");
8 q& F6 a, ?+ V' E! N- Z& P0 m[color=rgba(0, 0, 0, 0.749019607843137)] PostOrder(root);
' Z j# ^/ Y* ]# z* Y* O[color=rgba(0, 0, 0, 0.749019607843137)] printf("\n");
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[color=rgba(0, 0, 0, 0.749019607843137)] return 0;
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: u( s: Q d7 p[color=rgba(0, 0, 0, 0.749019607843137)]18
W5 x1 l R6 D$ T) T3 Z) {[color=rgba(0, 0, 0, 0.749019607843137)]19
5 e: f4 M' N1 I+ S r[color=rgba(0, 0, 0, 0.749019607843137)]20
# c! d* `9 t7 q" E2 D$ }4 `[color=rgba(0, 0, 0, 0.749019607843137)]21
" l( j' t" N2 X5 ^% w$ c* w1 Z[color=rgba(0, 0, 0, 0.749019607843137)]22
, T/ D7 ^/ `1 _% H( [ S) a$ w[color=rgba(0, 0, 0, 0.749019607843137)]23
/ w" Q4 a+ q( _5 r& G0 {[color=rgba(0, 0, 0, 0.749019607843137)]24; e1 i; ^: A! h- T* y1 ]
[color=rgba(0, 0, 0, 0.749019607843137)]25
: ~: p D4 h3 S6 ~ f3 q[color=rgba(0, 0, 0, 0.749019607843137)]26
5 k" Z* ]' u& N/ j/ b4 C[color=rgba(0, 0, 0, 0.749019607843137)]27& a `- M5 M* a; n, j
[color=rgba(0, 0, 0, 0.749019607843137)]28( r% f T! |7 u$ p) h' p l5 F
[color=rgba(0, 0, 0, 0.749019607843137)]29
5 f4 z' T$ E, v) I. f2 Q9 Z[color=rgba(0, 0, 0, 0.749019607843137)]30' [# B8 z) C M }% t
[color=rgba(0, 0, 0, 0.749019607843137)]31
6 L# c* y! G U6 k2 Z/ P) V[color=rgba(0, 0, 0, 0.749019607843137)]32% Z( Z! t. M3 t8 o
[color=rgba(0, 0, 0, 0.749019607843137)]33
8 U; ~, c- a# Z5 v Z[color=rgba(0, 0, 0, 0.749019607843137)]34
/ p$ u1 G* G' I$ J" H1 a6 I[color=rgba(0, 0, 0, 0.749019607843137)]35! x4 ?+ \1 H9 Q8 x* `: D
[color=rgba(0, 0, 0, 0.749019607843137)]36
# k: r4 F$ F2 j# I' o4 p[color=rgba(0, 0, 0, 0.749019607843137)]37
3 u) n& d0 k5 t R& {[color=rgba(0, 0, 0, 0.749019607843137)]38
* Z1 b4 @ ^: x# x[color=rgba(0, 0, 0, 0.749019607843137)]39
4 u! |0 F6 e: g; r) [* v[color=rgba(0, 0, 0, 0.749019607843137)]40/ t. d6 v% b# o" Z
[color=rgba(0, 0, 0, 0.749019607843137)]419 U. E- p. R1 s& C4 M% ]
[color=rgba(0, 0, 0, 0.749019607843137)]424 s2 R& p9 `5 P Z3 g& @8 K9 }
[color=rgba(0, 0, 0, 0.749019607843137)]43
! R* i( r% N, h# P9 X5 D; V[color=rgba(0, 0, 0, 0.749019607843137)]44
" p2 a; x0 K U& ][color=rgba(0, 0, 0, 0.749019607843137)]45; m8 ?1 F, d2 E
[color=rgba(0, 0, 0, 0.749019607843137)]46
" P* Z3 D+ u0 ?[color=rgba(0, 0, 0, 0.749019607843137)]47
( ~: o. S# c) s+ n* k2 F9 Y+ }[color=rgba(0, 0, 0, 0.749019607843137)]484 J: U ]2 F/ H% Z2 X
[color=rgba(0, 0, 0, 0.749019607843137)]49) }- s# D) h: J. ?
[color=rgba(0, 0, 0, 0.749019607843137)]50
, T0 D C1 J7 y[color=rgba(0, 0, 0, 0.749019607843137)]51$ G3 n Z. Q) S, G1 v/ L" o
[color=rgba(0, 0, 0, 0.749019607843137)]52
$ e* e" a, _, Z, l5 d7 X( q[color=rgba(0, 0, 0, 0.749019607843137)]53
3 l+ Y. {% D; ^[color=rgba(0, 0, 0, 0.749019607843137)]54# J) p: v% g, R& c. E% q
[color=rgba(0, 0, 0, 0.749019607843137)]55, O2 Q- J1 Y4 M5 b' Y$ _2 e
[color=rgba(0, 0, 0, 0.749019607843137)]569 U9 d7 C/ G* M* g' x
[color=rgba(0, 0, 0, 0.749019607843137)]57
# F$ d+ X$ Y8 O[color=rgba(0, 0, 0, 0.749019607843137)]58
& l" C* \# X, n4 M3 G[color=rgba(0, 0, 0, 0.749019607843137)]599 O: Q5 j8 x6 ~
[color=rgba(0, 0, 0, 0.749019607843137)]60/ B* ^' a1 O; e$ R; o G( @9 j
[color=rgba(0, 0, 0, 0.749019607843137)]61
2 ~ M- C1 i+ k% M {8 K* u[color=rgba(0, 0, 0, 0.749019607843137)]62; S5 R. j. k7 c, `' j* I
[color=rgba(0, 0, 0, 0.749019607843137)]63, z! L2 H, `! l/ [8 v1 C) S
[color=rgba(0, 0, 0, 0.749019607843137)]64) @, O0 s1 n4 y& z4 b% G
[color=rgba(0, 0, 0, 0.749019607843137)]65
+ n! }/ s. n( h% h/ N) R[color=rgba(0, 0, 0, 0.749019607843137)]66
9 R L5 H& `6 C3 z/ s2 b3 M" y[color=rgba(0, 0, 0, 0.749019607843137)]67$ f9 m% s9 \# r, P: O: n# w. ~4 Q
[color=rgba(0, 0, 0, 0.749019607843137)]68
/ F; \* Y! f6 _2 h5 r[color=rgba(0, 0, 0, 0.749019607843137)]69% y r! }( Y; ?5 Y
[color=rgba(0, 0, 0, 0.749019607843137)]70) L1 C+ a0 u% \; n; I, ?
[color=rgba(0, 0, 0, 0.749019607843137)]718 F8 X& B- ]1 A. a2 W
[color=rgba(0, 0, 0, 0.749019607843137)]726 h( j W$ q* s# ~
[color=rgba(0, 0, 0, 0.749019607843137)]730 g; S, K# o. L9 @ n/ Y
[color=rgba(0, 0, 0, 0.749019607843137)]74
9 G+ ?; Z0 x1 Y5 W; I[color=rgba(0, 0, 0, 0.749019607843137)]75- s# x5 t& E2 Z/ e
[color=rgba(0, 0, 0, 0.749019607843137)]76& p/ [3 a. y2 P& O
[color=rgba(0, 0, 0, 0.749019607843137)]776 F' t. V6 I" d/ ]: f
[color=rgba(0, 0, 0, 0.749019607843137)]78- z9 i- E7 m, b
[color=rgba(0, 0, 0, 0.749019607843137)]795 b2 `5 p4 O% {9 m$ i; o
[color=rgba(0, 0, 0, 0.749019607843137)]80
: r( H7 i" W2 e8 l[color=rgba(0, 0, 0, 0.749019607843137)]81
$ F; F4 J$ y' h[color=rgba(0, 0, 0, 0.749019607843137)]82, U4 R3 @; }, Q- @, H! U3 o
[color=rgba(0, 0, 0, 0.749019607843137)]83; n% u$ E# q% q4 X: ^" z) H
[color=rgba(0, 0, 0, 0.749019607843137)]840 A, p' y. r R/ a/ F
[color=rgba(0, 0, 0, 0.749019607843137)]85
1 w4 e( X6 t; B* A" O3 h9 {[color=rgba(0, 0, 0, 0.749019607843137)]86+ i( ]* K6 L- \. H
[color=rgba(0, 0, 0, 0.749019607843137)]87" F. I9 G* i0 Y8 w+ Z4 J- F
[color=rgba(0, 0, 0, 0.749019607843137)]88* Q+ z. O) t+ @" W( X. S4 P! v. K
[color=rgba(0, 0, 0, 0.749019607843137)]89
$ w6 W& L1 n& M( u+ x4 @ S: d[color=rgba(0, 0, 0, 0.749019607843137)]90( ^: C$ Q1 J4 ~
[color=rgba(0, 0, 0, 0.749019607843137)]917 _% C m# [3 S& a% ~
[color=rgba(0, 0, 0, 0.749019607843137)]92
% H, D- V& C' X W! [& |[color=rgba(0, 0, 0, 0.749019607843137)]93
' p2 \# B: b2 {0 u7 I[color=rgba(0, 0, 0, 0.749019607843137)]94
9 z0 V& J3 s& r9 x T9 V9 R[color=rgba(0, 0, 0, 0.749019607843137)]95
4 r6 @! f3 r2 Z- K7 ?- W[color=rgba(0, 0, 0, 0.749019607843137)]964 q' f) q3 {+ u9 A! p8 v
[color=rgba(0, 0, 0, 0.749019607843137)]97
Z6 p1 j# W2 A[color=rgba(0, 0, 0, 0.749019607843137)]98# i3 i" v5 u x8 C( c
[color=rgba(0, 0, 0, 0.749019607843137)]99" E* T' w$ ^. Z9 Z& U' T
[color=rgba(0, 0, 0, 0.749019607843137)]100
u M( S( c# V9 }[color=rgba(0, 0, 0, 0.749019607843137)]101* R" s! r6 F- I! H9 M! F2 N% \
[color=rgba(0, 0, 0, 0.749019607843137)]102" p/ D" F" ~; D* B) |* I
[color=rgba(0, 0, 0, 0.749019607843137)]1037 \# l' w) D; x6 l1 O
[color=rgba(0, 0, 0, 0.749019607843137)]104
& `5 j1 @' a8 B9 W+ \, V[color=rgba(0, 0, 0, 0.749019607843137)]105
8 b i* i3 n% l0 \# W[color=rgba(0, 0, 0, 0.749019607843137)]106
7 c5 C. S/ F* G3 R4 t ]* F. @/ L+ [5 s) {! w[color=rgba(0, 0, 0, 0.749019607843137)]107
0 r" i- l" G( }- ~3 f- I4 T[color=rgba(0, 0, 0, 0.749019607843137)]108
1 g! ~. o* h0 D[color=rgba(0, 0, 0, 0.749019607843137)]109- `- h2 h! t, a6 Y
[color=rgba(0, 0, 0, 0.749019607843137)]110/ y0 Y2 {; k3 L
[color=rgba(0, 0, 0, 0.749019607843137)]1117 E3 c U5 E# d6 C' C- {. D. e
[color=rgba(0, 0, 0, 0.749019607843137)]测试结果：# u# L I# W. B$ L* K) p) p+ A
[color=rgba(0, 0, 0, 0.749019607843137)]
. k5 C g2 e( X; P) ^ f% s" s$ u$ T9 w5 U- C
[color=rgba(0, 0, 0, 0.749019607843137)]9 H: B; j3 K" h+ L. R. _! O
1 l" F9 E ]" [/ c K
[color=rgba(0, 0, 0, 0.749019607843137)]2.2计算二叉树的大小, ^/ G( W' f ]0 Q0 V3 k
[color=rgba(0, 0, 0, 0.749019607843137)]时间复杂度为O(N)
2 V) z+ Z0 R7 B[color=rgba(0, 0, 0, 0.749019607843137)]1 ]: H' O& G$ @% G! i" n
' z# k7 A+ _& G; g[color=rgba(0, 0, 0, 0.749019607843137)]//二叉树的大小; b0 d* f7 O ]( a: z
[color=rgba(0, 0, 0, 0.749019607843137)]//法一：全局变量
2 o& ]5 t3 J$ Q3 K. L) u- H+ _[color=rgba(0, 0, 0, 0.749019607843137)]//int count = 0;
3 \. } l1 Y p7 |: _[color=rgba(0, 0, 0, 0.749019607843137)]//void TreeSize(BTNode* root)
: B4 P' [$ D2 A[color=rgba(0, 0, 0, 0.749019607843137)]//{
; U; |, p$ y2 D i) c+ h[color=rgba(0, 0, 0, 0.749019607843137)]// if (root == NULL)
0 W2 {% ]7 o+ d0 G7 ~[color=rgba(0, 0, 0, 0.749019607843137)]// {
4 Z8 L6 X0 c R: A[color=rgba(0, 0, 0, 0.749019607843137)]// return;
$ M# ^9 J+ f6 f+ e+ f% X[color=rgba(0, 0, 0, 0.749019607843137)]// }
+ ?5 ^& C* c0 V* H: G# X) g2 T[color=rgba(0, 0, 0, 0.749019607843137)]// count++;4 o) A8 s# K: N; J [" i# P$ ?
[color=rgba(0, 0, 0, 0.749019607843137)]// TreeSize(root->left);
. z- A% p9 T+ P5 i' Y. u[color=rgba(0, 0, 0, 0.749019607843137)]// TreeSize(root->right);, ?' U! F/ I8 e! Y* P
[color=rgba(0, 0, 0, 0.749019607843137)]//
6 k m @6 o$ {. D, h8 s6 W/ U; h[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- s/ Q1 ?8 M0 Q/ j" O8 J
[color=rgba(0, 0, 0, 0.749019607843137)]//}
* l2 A7 n2 C7 Q* a[color=rgba(0, 0, 0, 0.749019607843137)]//法二：子问题思路：分而治之# I3 v! y ^! \. ?. t( ^ K
[color=rgba(0, 0, 0, 0.749019607843137)]int TreeSize(BTNode* root)/ _; H9 N! {- s6 o1 Q) ]
[color=rgba(0, 0, 0, 0.749019607843137)]{% K+ L$ i1 r- x! l( r9 V% ^- d* r- _
[color=rgba(0, 0, 0, 0.749019607843137)] return root == NULL ? 0 : TreeSize(root->left) + TreeSize(root->right) + 1;; E+ Z# H- r8 w' x) r( b" X
[color=rgba(0, 0, 0, 0.749019607843137)]}
% w( F: [& O" L! R# ^' r* _2 w[color=rgba(0, 0, 0, 0.749019607843137)]1
7 D/ B. S% f/ _+ K4 z* P[color=rgba(0, 0, 0, 0.749019607843137)]2+ p. P: p- }8 v- d4 F
[color=rgba(0, 0, 0, 0.749019607843137)]3
( k) g8 \+ g1 O. r) H8 x[color=rgba(0, 0, 0, 0.749019607843137)]4
- Y9 Z7 c& q4 \/ o d[color=rgba(0, 0, 0, 0.749019607843137)]5( `+ J& o& y1 O
[color=rgba(0, 0, 0, 0.749019607843137)]6
) X* ^6 Y% k3 f! g& E9 M" u2 g[color=rgba(0, 0, 0, 0.749019607843137)]7
, p# j0 A2 g" O0 D! D7 m0 J8 M; J[color=rgba(0, 0, 0, 0.749019607843137)]8" }+ S" M1 v& T2 `! ]1 f
[color=rgba(0, 0, 0, 0.749019607843137)]9
& B: _$ r1 _1 X; C, U7 Z# }[color=rgba(0, 0, 0, 0.749019607843137)]10' r$ o0 |5 c8 a/ f3 x
[color=rgba(0, 0, 0, 0.749019607843137)]111 q" q U- t( W0 C, ?
[color=rgba(0, 0, 0, 0.749019607843137)]12
" R/ m# P' z, f! s0 B7 i[color=rgba(0, 0, 0, 0.749019607843137)]132 B8 Z( V+ P' @; S
[color=rgba(0, 0, 0, 0.749019607843137)]14
2 W [* G3 L/ r+ P5 i9 B7 f9 h5 I[color=rgba(0, 0, 0, 0.749019607843137)]15
1 ~; h: N6 }% l) v+ P, s[color=rgba(0, 0, 0, 0.749019607843137)]16) ]% N5 @4 @% ?/ p" Y$ M
[color=rgba(0, 0, 0, 0.749019607843137)]17
# c; \( p9 H1 I6 Q5 R9 d! a" F[color=rgba(0, 0, 0, 0.749019607843137)]187 y7 B, |" j3 J
[color=rgba(0, 0, 0, 0.749019607843137)]19' J+ C/ _0 E/ G& w, G
[color=rgba(0, 0, 0, 0.749019607843137)]20
" k# C, j; s% Y% W+ V4 }. D[color=rgba(0, 0, 0, 0.749019607843137)]2.3计算二叉树叶子结点的个数
1 X) n$ M& V# a" x; t9 N[color=rgba(0, 0, 0, 0.749019607843137)]//计算二叉树叶子结点的个数 d+ c1 r* l$ k1 v0 }$ }
[color=rgba(0, 0, 0, 0.749019607843137)]int TreeLeafSize(BTNode* root)
7 J4 I6 j# ]. k) f" D[color=rgba(0, 0, 0, 0.749019607843137)]{. Y+ z6 U1 t# U4 T; q
[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( W/ B6 K0 U9 j0 v& V1 l
[color=rgba(0, 0, 0, 0.749019607843137)] {7 s" ^; [5 D1 u1 r; F
[color=rgba(0, 0, 0, 0.749019607843137)] return 0;
& @, y% a3 ^: |/ D$ i( _$ b[color=rgba(0, 0, 0, 0.749019607843137)] }4 q- g- n% b- X4 B* `
[color=rgba(0, 0, 0, 0.749019607843137)] //叶子结点的特征就是左右子树为空' B7 L* l) w- c0 t, L2 k0 `
[color=rgba(0, 0, 0, 0.749019607843137)] if (root->left == NULL && root->right == NULL)$ C) O; F: _+ E2 Z; y4 ~, P2 @
[color=rgba(0, 0, 0, 0.749019607843137)] {
+ d- J7 v5 a+ f5 I+ m9 V[color=rgba(0, 0, 0, 0.749019607843137)] return 1;
+ K3 B7 i6 {5 D& z[color=rgba(0, 0, 0, 0.749019607843137)] }
' y; t+ ~1 f& G[color=rgba(0, 0, 0, 0.749019607843137)] return TreeLeafSize(root->left) + TreeLeafSize(root->right);; O6 E! J( S) r8 F1 ~0 j
[color=rgba(0, 0, 0, 0.749019607843137)]}
4 B3 L* }; x" e0 I[color=rgba(0, 0, 0, 0.749019607843137)]1
& y, ?6 }9 ]- F% R; ?[color=rgba(0, 0, 0, 0.749019607843137)]2
0 Y# j5 }3 |( I+ T, J2 p7 K6 r. a[color=rgba(0, 0, 0, 0.749019607843137)]36 U2 k L3 p0 |( L8 k3 N! \6 c/ g
[color=rgba(0, 0, 0, 0.749019607843137)]4
: G) t; i6 X) Q) v [[color=rgba(0, 0, 0, 0.749019607843137)]5- V, |( q3 G5 G0 l! I6 m
[color=rgba(0, 0, 0, 0.749019607843137)]6( K6 |! L. d" Y4 J- ?! U. Z% G8 F
[color=rgba(0, 0, 0, 0.749019607843137)]7. Z. }; d! q6 X' t$ p8 h
[color=rgba(0, 0, 0, 0.749019607843137)]80 t7 k/ _; }* B
[color=rgba(0, 0, 0, 0.749019607843137)]9; I' G8 O1 }' [8 E
[color=rgba(0, 0, 0, 0.749019607843137)]10& f9 z c3 _/ X* A& e; Q
[color=rgba(0, 0, 0, 0.749019607843137)]11
% J- h7 D" J+ ?+ P[color=rgba(0, 0, 0, 0.749019607843137)]12
+ q" Y M& D+ g' t! F[color=rgba(0, 0, 0, 0.749019607843137)]13# L. E7 h/ u- ]8 O) _- l
[color=rgba(0, 0, 0, 0.749019607843137)]140 k/ @, G' l y* o/ r0 r- K
[color=rgba(0, 0, 0, 0.749019607843137)]2.4计算二叉树的高度
* @: O. [ w% W d! H2 z9 H[color=rgba(0, 0, 0, 0.749019607843137)]int TreeHeight(BTNode* root)
6 ]% J9 [' L# F; F[color=rgba(0, 0, 0, 0.749019607843137)]{
9 k* `5 L% [5 F2 G[color=rgba(0, 0, 0, 0.749019607843137)] //空树高度为0 u3 R* ?" |4 j- V0 X0 O7 H" i
[color=rgba(0, 0, 0, 0.749019607843137)] if (root == NULL)' {2 V' b" j- P1 o$ I
[color=rgba(0, 0, 0, 0.749019607843137)] {
K& X1 w! I# ]% `9 Y$ \0 ~, W[color=rgba(0, 0, 0, 0.749019607843137)] return 0;9 N, ? A h; w) w5 S
[color=rgba(0, 0, 0, 0.749019607843137)] }% N( ~* q: N, R& k, u/ q: D' @5 K/ _
[color=rgba(0, 0, 0, 0.749019607843137)] //树的高度是较高的那棵子树6 D+ v6 W3 L" E0 [+ ^" n; [* J
[color=rgba(0, 0, 0, 0.749019607843137)] int lh = TreeHeight(root->left);//左子树的高度7 E# T+ G, P+ v
[color=rgba(0, 0, 0, 0.749019607843137)] int rh = TreeHeight(root->right);//右子树的高度* _8 p0 ]; x0 V9 y' I# {0 E+ h( T0 b
[color=rgba(0, 0, 0, 0.749019607843137)]
4 z% |* @0 `' ]( d9 F: U% ?- m O
3 {) G( D% Q0 V9 a[color=rgba(0, 0, 0, 0.749019607843137)] return lh > rh ? lh + 1 : rh + 1;/ t8 w+ Y5 H- `: P7 ]. [; P. R7 B
[color=rgba(0, 0, 0, 0.749019607843137)]}2 i' b4 L. F Z* t9 g* F
[color=rgba(0, 0, 0, 0.749019607843137)]1
6 z, H2 |8 E" b/ Y" r( D; D$ ?[color=rgba(0, 0, 0, 0.749019607843137)]2
! K0 a7 J3 I: r/ d[color=rgba(0, 0, 0, 0.749019607843137)]3
$ Q. _9 c4 w9 V4 w* U[color=rgba(0, 0, 0, 0.749019607843137)]4
" a. [1 c) x8 a/ Q j[color=rgba(0, 0, 0, 0.749019607843137)]5
+ e- }& V" w; A% y% T) ~/ W) Z' P[color=rgba(0, 0, 0, 0.749019607843137)]6
" v* ~/ E& u' g7 E+ i( R2 ?[color=rgba(0, 0, 0, 0.749019607843137)]7; }! s! h$ h1 z9 B+ j* a) A: \4 i# K. v
[color=rgba(0, 0, 0, 0.749019607843137)]8
, ]$ p. E+ y' h# g \+ B3 I0 d[color=rgba(0, 0, 0, 0.749019607843137)]90 p: B% [2 h% b- v: C E. P. _$ ^
[color=rgba(0, 0, 0, 0.749019607843137)]10' B& T1 i- ~( X7 c9 K
[color=rgba(0, 0, 0, 0.749019607843137)]119 d# `, \# s1 Q1 C4 j5 H! | ?
[color=rgba(0, 0, 0, 0.749019607843137)]12/ ^9 V6 T0 i2 P' K' E+ _
[color=rgba(0, 0, 0, 0.749019607843137)]13# m" w! _& p' N: m
[color=rgba(0, 0, 0, 0.749019607843137)]2.5计算第K层结点的个数6 @) I' q3 Z' z$ ?2 q7 L) d
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[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。。。; f! V: a4 o9 z' Z4 A3 w/ C1 |
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b5 M" Q$ V# K' x- Y. m[color=rgba(0, 0, 0, 0.749019607843137)]//计算第K层结点的个数
5 G& V) s% @2 o$ M: n5 s0 G2 M[color=rgba(0, 0, 0, 0.749019607843137)]int TreeLevel(BTNode* root,int K)1 N1 f5 D0 ]. x4 X& u
[color=rgba(0, 0, 0, 0.749019607843137)]{
8 T9 g8 W+ F7 o- Q1 f[color=rgba(0, 0, 0, 0.749019607843137)] assert(K > 0);
. D' ^6 W. b( B! w4 }[color=rgba(0, 0, 0, 0.749019607843137)] if (root == NULL)' t* Y3 _$ i6 _% L1 o1 U
[color=rgba(0, 0, 0, 0.749019607843137)] {
8 U2 v t0 P) f4 J0 C- K[color=rgba(0, 0, 0, 0.749019607843137)] return 0;0 Z- @- R; H7 A4 H. v
[color=rgba(0, 0, 0, 0.749019607843137)] }& ]; R, _- P! Z; k7 ?" i3 u
[color=rgba(0, 0, 0, 0.749019607843137)] //如果是第一层(递归出口) e6 `3 Z5 n! N; ?3 P
[color=rgba(0, 0, 0, 0.749019607843137)] if (K == 1)
5 X$ ?9 ?! Z7 O N# j3 e2 Y[color=rgba(0, 0, 0, 0.749019607843137)] {
1 [" x% R ^3 a[color=rgba(0, 0, 0, 0.749019607843137)] return 1;0 K. |8 c: n( Y: y, p
[color=rgba(0, 0, 0, 0.749019607843137)] }$ f! P* [- ~- u7 E0 C9 |- r2 j4 E# B
[color=rgba(0, 0, 0, 0.749019607843137)] //转换成子树的第K-1层
9 c M, m! ?" X# P# n[color=rgba(0, 0, 0, 0.749019607843137)] return TreeLevel(root->left,K-1) + TreeLevel(root->right,K-1);0 N6 l# w$ v e' U
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" ]- _- T3 ~! w0 l[color=rgba(0, 0, 0, 0.749019607843137)]10% ~8 v$ X3 E( s5 i1 d. C) Y, b) I' X
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[color=rgba(0, 0, 0, 0.749019607843137)]160 {! A- C! k2 Y7 ]" w& W* u, F
[color=rgba(0, 0, 0, 0.749019607843137)]2.6二叉树查找8 [) b% h c4 h- b) @0 j
[color=rgba(0, 0, 0, 0.749019607843137)]//二叉树查找
& M# A- x. k* L" \& C& x0 w[color=rgba(0, 0, 0, 0.749019607843137)]BTNode* TreeFind(BTNode* root, BTDataType data): G! W0 k3 G" ^8 x8 P
[color=rgba(0, 0, 0, 0.749019607843137)]{* t3 |. T d( Y7 h# l
[color=rgba(0, 0, 0, 0.749019607843137)] if (root == NULL)
1 J- D# Y" Y( s3 V3 K9 v) g[color=rgba(0, 0, 0, 0.749019607843137)] {
! i# {% g* b5 {; I[color=rgba(0, 0, 0, 0.749019607843137)] return NULL;
. T9 }5 `; z2 o) G[color=rgba(0, 0, 0, 0.749019607843137)] }
$ w- z; i6 a z+ }$ I' ]3 \: k[color=rgba(0, 0, 0, 0.749019607843137)] if (root->data == data), S- X6 g, j. ^2 y( z3 }+ Y
[color=rgba(0, 0, 0, 0.749019607843137)] {
* S2 a& K, j+ C- Q/ @[color=rgba(0, 0, 0, 0.749019607843137)] return root;4 s* r( u* H: g7 c$ g# ^
[color=rgba(0, 0, 0, 0.749019607843137)] }9 r, B$ U9 |8 r. _+ W, A6 h0 H
[color=rgba(0, 0, 0, 0.749019607843137)] //先查找左子树
6 r4 z |7 }* }/ j4 }1 R[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* lret = TreeFind(root->left, data);
8 A1 {, S0 U& B8 n8 U[color=rgba(0, 0, 0, 0.749019607843137)] if (lret), T& L; w+ I( Y! d
[color=rgba(0, 0, 0, 0.749019607843137)] return lret;
6 ^! [* o0 T3 `7 N' q[color=rgba(0, 0, 0, 0.749019607843137)] //再查找右子树
7 e5 a! {5 \0 v/ g. s7 T[color=rgba(0, 0, 0, 0.749019607843137)] BTNode* rret = TreeFind(root->right, data);% k5 f) y, T" i; W6 N0 V/ b
[color=rgba(0, 0, 0, 0.749019607843137)] if (rret) k, I1 J& a3 S! Y0 w" i1 p( j5 w
[color=rgba(0, 0, 0, 0.749019607843137)] return rret;
: G. ]% ~0 _5 O' g[color=rgba(0, 0, 0, 0.749019607843137)] return NULL;
" q* h$ t! V# q5 C# `+ _3 u/ _[color=rgba(0, 0, 0, 0.749019607843137)]}
! N$ b" A8 n( D/ D4 o1 n[color=rgba(0, 0, 0, 0.749019607843137)]————————————————8 ^( S' Y, I/ r* |
[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。; S. Y& R+ ^2 Q ]6 S A
[color=rgba(0, 0, 0, 0.749019607843137)]原文链接：https://blog.csdn.net/weixin_63449996/article/details/1268412120 E5 A$ ^! m5 C- h) q# n" V7 V
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