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// Date :2016.07.22
// Author :qytao ///************************************************************************************ 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.** OJ's Binary Tree Serialization:** The serialization of a binary tree follows a level order traversal, where '#' signifies* a path terminator where no node exists below.** Here's an example:** 1* / \* 2 3* /* 4* \* 5** The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".************************************************************************************/ //we have two ways to solve this problems
//first rescure 递归方法,与判断是否是同一颗树非常相似。class Solution {public: bool isSymmetric(TreeNode *p, TreeNode *q) { if (p == NULL && q == NULL) return true; if (p == NULL || q == NULL) return false; return (p->val == q->val) && isSymmetric(p->left, q->right) && isSymmetric(p->right, q->left); } bool isSymmetric(TreeNode* root) { if (!root) return true; return isSymmetric(root->left, root->right); }};//the next solutions iterative solutionclass Solution {public: bool isSymmetric(TreeNode* root) { if (!root) return true; stack sk; sk.push(root->left); sk.push(root->right); TreeNode* pA, *pB; while (!sk.empty()) { pA = sk.top(); sk.pop(); pB = sk.top(); sk.pop(); if (!pA && !pB) continue; if (!pA || !pB) return false; if (pA->val != pB->val) return false; sk.push(pA->left); sk.push(pB->right); sk.push(pA->right); sk.push(pB->left); } return true; }}; 转载地址：http://avdoi.baihongyu.com/