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
题解：递归的解法，重新写一个递归函数bool issysmem(TreeNode *lc,TreeNode &rc)，函数输入分别输入待比较是否对称的两个树的树根。首先比较这两棵树根的值是否相等，然后递归的比较lc的左子和rc的右子，以及lc的右子和rc的左子是否对称。代码如下:
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool isSymmetric(TreeNode *root) { 13 if(root == NULL) 14 return true; 15 return issysmem(root->left,root->right); 16 } 17 bool issysmem(TreeNode *lc,TreeNode *rc){ 18 if((!lc && rc) || (lc && !rc)) 19 return false; 20 if(lc == NULL && rc == NULL) 21 return true; 22 if(lc->val != rc->val) 23 return false; 24 return issysmem(lc->left,rc->right) && issysmem(lc->right,rc->left); 25 } 26 };
【leetcode】Symmetric Tree,布布扣,bubuko.com
原文：http://www.cnblogs.com/sunshineatnoon/p/3779044.html