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题目：
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1 / 2 2 / \ / 3 4 4 3
But the following is not:
1 / 2 2 \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
代码：
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isSymmetric(TreeNode* root) { if (!root) return true; stack<TreeNode *> staL, staR; staL.push(root->left); staR.push(root->right); while ( !staL.empty() && !staR.empty() ) { TreeNode *tmpL = staL.top(); staL.pop(); TreeNode *tmpR = staR.top(); staR.pop(); if ( !tmpL && !tmpR ) continue; if ( !tmpL || !tmpR ) return false; if ( tmpL->val != tmpR->val ) return false; staL.push(tmpL->right); staL.push(tmpL->left); staR.push(tmpR->left); staR.push(tmpR->right); } return staL.empty() && staR.empty(); } };
tips：
深搜思想。设立两个栈：左栈和右栈。
从根节点开始：左子树用左栈遍历（node->left->right）；右子树用右栈遍历(node->right->left)。
这样就可以转化为Same Tree这道题了（http://www.cnblogs.com/xbf9xbf/p/4505032.html）
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4505116.html
