LeetCode Symmetric Tree

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.

1. 递归代码

/**
 * 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) {
    	if(root == NULL)
    		return true;
    	else 
    		return isMirror(root->left, root->right);        
    }
private:
	bool isMirror(TreeNode *p1, TreeNode *p2) {
		if(p1 == NULL && p2 == NULL)
			return true;
		else if(p1 == NULL || p2 == NULL)
			return false;
		else if(p1->val != p2->val)
			return false;
		else
			return (isMirror(p1->left, p2->right) && isMirror(p1->right, p2->left));
	}
};

2. 迭代代码

/**
 * 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) {
    	if(root == NULL)
    		return true;
    	queue<TreeNode*> left_que, right_que;
    	left_que.push(root->left);
    	right_que.push(root->right);
    	while(!left_que.empty() && !right_que.empty()) {
    		TreeNode *cur_left = left_que.front();
    		TreeNode *cur_right = right_que.front();
    		left_que.pop();
    		right_que.pop();
    		if(cur_left == NULL && cur_right == NULL)
    		    continue;
    		else if(cur_left == NULL || cur_right == NULL)
    			return false;
    		else if(cur_left->val != cur_right->val)
    			return false;
    		left_que.push(cur_left->left);
    		left_que.push(cur_left->right);
    		right_que.push(cur_right->right);
    		right_que.push(cur_right->left);	
    	}	       
    }
};