剑指offer(18)—二叉树的镜像

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

思路

递归 or 非递归

代码

递归

/*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;        //交换左右子结点        TreeNode* ptmp;        ptmp = pRoot->left;        pRoot->left = pRoot->right;        pRoot->right = ptmp;        Mirror(pRoot->left);    //对左子树进行镜像操作        Mirror(pRoot->right);   //对右子树进行镜像操作        return;    }};

非递归

/*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) {        //栈的非递归        if(pRoot == NULL)            return;        stack<TreeNode *> st;        st.push(pRoot);        TreeNode *ptmp, *ptop;        while(st.size() > 0){            ptop = st.top();            st.pop();            ptmp = ptop->left;            ptop->left = ptop->right;            ptop->right = ptmp;            if(ptop->left)                st.push(ptop->left);            if(ptop->right)                st.push(ptop->right);        }        return;    }};