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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
这道题目可以用迭代的方式做，也可以用递归的方式做。遇见和树相关的题目，第一反应是递归，所以我的解法是递归。
代码奉上：
class Solution { public: bool IsSymmetric(TreeNode *node1, TreeNode *node2) { if (node1 == NULL&&node2 == NULL) return true; if (node1 == NULL||node2 == NULL) return false; bool result1, result2; if (node1->val != node2->val) return false; else { result1 = IsSymmetric(node1->left, node2->right); result2 = IsSymmetric(node1->right, node2->left); } return result1&&result2; } bool isSymmetric(TreeNode *root) { if (root == NULL || root->left == NULL&&root->right == NULL) return true; bool result = IsSymmetric(root->left, root->right); return result; } };
看了相关人的解法，有的人直接用迭代的方式解出来了，感觉很受启发
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isSymmetric(TreeNode *root) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. if(root == NULL)return true; queue<TreeNode*> qleft, qright; if(root->left)qleft.push(root->left); if(root->right)qright.push(root->right); while(qleft.empty() == false && qright.empty() == false) { TreeNode *ql = qleft.front(); TreeNode *qr = qright.front(); qleft.pop(); qright.pop(); if(ql->val == qr->val) { if(ql->left && qr->right) { qleft.push(ql->left); qright.push(qr->right); } else if(ql->left || qr->right) return false; if(qr->left && ql->right) { qleft.push(qr->left); qright.push(ql->right); } else if(qr->left || ql->right) return false; } else return false; } if(qleft.empty() && qright.empty()) return true; else return false; } };
转自http://www.cnblogs.com/TenosDoIt/p/3440729.html
HappyLeetcode43:Symmetric Tree
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原文地址：http://www.cnblogs.com/chengxuyuanxiaowang/p/4219870.html
