18、剑指offer--二叉树的镜像

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

 

解题思路：

1）若左右子结点不为空则交换左右子结点

2）对交换后的左子结点递归处理

3）对交换后的右子结点递归处理

#include <iostream>
#include <malloc.h>
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;
        if(pRoot->left == NULL && pRoot->right == NULL)
            return;
        //交换左右子结点
        TreeNode *temp = pRoot->left;
        pRoot->left = pRoot->right;
        pRoot->right = temp;
        //交换后的左子节点不为空，递归
        if(pRoot->left != NULL)
            Mirror(pRoot->left);
        //交换后的右子节点不为空，递归
        if(pRoot->right != NULL)
            Mirror(pRoot->right);
    }
};
//先序建立二叉树
TreeNode *CreateBiTree(){
    int ch;

    TreeNode *T;
    cin>>ch;
    if(ch== 0)T=NULL;
    else{
        T = (TreeNode *)malloc(sizeof(TreeNode));
        T->val = ch;
        cout<<"输入左子节点"<<endl;
        T->left = CreateBiTree();
        cout<<"输入右子节点"<<endl;
        T->right = CreateBiTree();
    }
    return T;//返回根节点
}
//先序遍历二叉树
void PreOrderTraverse(TreeNode *T){
    if(T)
    {
       cout<<T->val<<" ";
       PreOrderTraverse(T->left);
       PreOrderTraverse(T->right);
    }
}
int main()
{
    TreeNode *T1;
    cout<<"输入根节点："<<endl;
    T1 = CreateBiTree();//建立
    cout<<"原二叉树先序遍历"<<endl;
    PreOrderTraverse(T1);//输出
    cout<<endl;
    Solution s;
    s.Mirror(T1);
    cout<<"镜像后的二叉树先序遍历"<<endl;
    PreOrderTraverse(T1);//输出
    return 0;
}