版权声明： https://blog.csdn.net/shuiyixin/article/details/83352947
一、前言
哈哈，今天就是程序员节啦，祝大家1024程序员节快乐。
今天要给大家分享的算法是交换二叉树是的左右孩子结点，比较简单，需要创建一个结点用来暂存左孩子结点，下面给大家送上代码。
二、题目
将下图用二叉树存入，并交换二叉树是的左右孩子结点。其中圆角矩形内为结点数据，旁边数字为结点编号，编号为0的结点为根节点，箭头指向的结点为箭尾的孩子结点。
三、代码
#define MAXQUEUESIZE 10
#include<iostream>
#include<malloc.h>
using namespace std;
typedef struct BiTNode {
int data;
int number;
struct BiTNode *lChild, *rChild, *parent;
}BiTNode, *BiTree;
int number = 0;
int yon = 0;
int yesOrNo[] = { 1,0,1,0,0,1,1,1,0,0,1,0,0,1,0,0 };
int numData[] = { 1,2,4,3,5,7,8,6 };
BiTree treeParent = NULL;
int OperationBiTree(BiTree &T) {
T = (BiTree)malloc(sizeof(BiTNode));
if (!T)
{
cout << "空间分配失败(Allocate space failure.)" << endl;
exit(OVERFLOW);
}
T->data = numData[number];
T->number = number;
number++;
T->lChild = NULL;
T->rChild = NULL;
T->parent = treeParent;
return 1;
}
int RecursionEstablishBiTree(BiTree &T) {
OperationBiTree(T);
treeParent = T;
if (yesOrNo[yon++])
RecursionEstablishBiTree(T->lChild);
treeParent = T;
if (yesOrNo[yon++])
RecursionEstablishBiTree(T->rChild);
return 1;
}
void VisitBiTree(BiTree &T) {
cout << "The number of present node is ：" << T->number << "; ";
cout << " data is ：" << T->data << ";\n ";
if (T->lChild)
cout << " has left child and the number is :" << T->lChild->number << ";\n";
if (T->rChild)
cout << " has right child,and the number is :" << T->rChild->number << ";\n";
cout << endl;
}
//Visit tree use the preorder technique.
void PreOrderVisitBiTree(BiTree T) {
if (T)
{
VisitBiTree(T);
PreOrderVisitBiTree(T->lChild);
PreOrderVisitBiTree(T->rChild);
}
}
int ExchangeChildNode(BiTree &T) {
BiTree p;
if (T)
{
ExchangeChildNode(T->lChild);
ExchangeChildNode(T->rChild);
p = T->lChild;
T->lChild = T->rChild;
T->rChild = p;
}
return 1;
}
void main() {
BiTree T;
RecursionEstablishBiTree(T);
cout << "************【Recursion of the binary tree just establish】************\n";
PreOrderVisitBiTree(T);
ExchangeChildNode(T);
cout << "*******【Recursion of the binary tree just exchange child node】*******\n";
PreOrderVisitBiTree(T);
}四、实现效果
创建效果如下，遍历访问测试：
对于第二颗树，只需要修改两个位置，树创建时候的两个数组：
int yesOrNo[] = { 1,1,1,0,0,1,0,0,1,0,0,1,1,0,0,1,0,0 };
int numData[] = { 1,2,4,8,9,5,3,6,7 };结果如下：
