判断对称树

题目描述

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}".

利用栈求解；
 

实现代码：

               /**
 * 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;
        stack<TreeNode*> s1,s2;
    
        s1.push(root->left);
        s2.push(root->right);
        while(!s1.empty()&&!s2.empty())
        {
           TreeNode* n1 = s1.top();  
            s1.pop();
           TreeNode* n2 = s2.top();  
            s2.pop();
           if (n1 == NULL && n2 == NULL)  
               continue;  
           if (n1 == NULL || n2 == NULL)  
               return false;  
           if (n1->val != n2->val)  
               return false;  
           s1.push(n1->left);  //s1,s2压入左右子树的顺序相反！！！
           s2.push(n2->right);  
           s1.push(n1->right);  
           s2.push(n2->left);  
        }
        return true;
    }
};