二叉树的镜像（数组，前后 遍历重建二叉树）

题目描述

操作给定的二叉树，将其变换为源二叉树的镜像。

输入描述:

二叉树的镜像定义：源二叉树 8/  \6   10/ \  / \

	5  7 9 11镜像二叉树8/  \10   6/ \  / \11 9 7  5

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;struct TreeNode
{int val;struct TreeNode *left;struct TreeNode *right;TreeNode(int x):val(x),left(NULL),right(NULL){}
};class Solution
{
public:void Mirror(TreeNode *pRoot){if(pRoot==NULL)return ;TreeNode *tmpNode;if(pRoot->left||pRoot->right){tmpNode=pRoot->left;pRoot->left=pRoot->right;pRoot->right=tmpNode;if(pRoot->left){Mirror(pRoot->left);}if(pRoot->right){Mirror(pRoot->right);}}}TreeNode* ConstructT(int *preorder,int *inorder,int len){TreeNode* pRoot = (TreeNode*)malloc(sizeof(TreeNode));int index =0;while(inorder[index]!=preorder[0]&&index<len){index++;}if(index==len){return NULL;}pRoot->val=preorder[0];pRoot->left=ConstructT(preorder+1,inorder,index);pRoot->right=ConstructT(preorder+index+1,inorder+index+1,len-index);return pRoot;}void PrintPreorder(TreeNode *pRoot){if(pRoot){cout<<pRoot->val;PrintPreorder(pRoot->left);PrintPreorder(pRoot->right);}}
};int main()
{Solution s;TreeNode* pRoot;int preorder[]={8,6,5,7,10,9,11};int inorder[]={5,6,7,8,9,10,11};int len=7;pRoot=s.ConstructT(preorder,inorder,len);s.PrintPreorder(pRoot);s.Mirrior(pRoot);s.PrintPreorder(pRoot);return 0;
}