!本题分为迭代版和递归版，递归版比较直接，按照树遍历即可。跌代版这里创建了两个队列，每遇到一对节点（初始为根节点的左右子）：
1.先对比这两个节点的值是否相等：不相等则结束循环，相等则进入2
2.把它们的左右子节点，交叉安插到自己的队列中，然后对队列里的front，作为一对节点，走1
!iterative
1 if(!root) return true; 2 queue<TreeNode*> queueStep; 3 queue<TreeNode*> queueStep2; 4 TreeNode *tn1,*tn2; 5 tn1 = root->left; 6 tn2 = root->right; 7 while(1){ 8 if(!tn1 && tn2) return false; 9 if(!tn2 && tn1) return false; 10 if(tn1 && tn2 && tn1->val != tn2->val) return false; 11 12 if(tn1){ 13 queueStep.push(tn1->left); 14 queueStep.push(tn1->right); 15 } 16 if(tn2){ 17 queueStep2.push(tn2->right); 18 queueStep2.push(tn2->left); 19 } 20 if(queueStep.size() == 0 && queueStep2.size() == 0) return true; 21 /*else if(queueStep.size() == 0) return false; 22 else if(queueStep2.size() == 0) return false;*/ 23 24 tn1 = queueStep.front(); 25 queueStep.pop(); 26 27 tn2 = queueStep2.front(); 28 queueStep2.pop(); 29 }
!recursive
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) return true; 14 return isSymmetricSmall(root->left,root->right); 15 } 16 17 bool isSymmetricSmall(TreeNode *left,TreeNode *right){ 18 if(left == NULL && right == NULL) return true; 19 else if(left == NULL) return false; 20 else if(right == NULL) return false; 21 22 if(left->val != right->val) return false; 23 24 if(isSymmetricSmall(left->right,right->left) && isSymmetricSmall(left->left,right->right)) 25 return true; 26 else 27 return false; 28 } 29 };