【LeetCode】101

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.

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}".

Solution 1: 递归，left对应right，left->left对应right->right，left->right对应right->left

/**
 * Definition for a binary tree node.
 * 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)return true;
        return isSymTree(root->left,root->right);
    }
    bool isSymTree(TreeNode *p,TreeNode *q){
        if(!isSameNode(p,q))
            return false;
        else if(!p&&!q)
            return true;
        else
            return isSymTree(p->left,q->right) && isSymTree(p->right,q->left);
    }
    bool isSameNode(TreeNode *p,TreeNode *q){
        if(!p&&!q)      //必需加上这个判断条件，否则若p、q为空下面的->val会运行时错误
            return true;
        else if((!p&&q)||(p&&!q)||(p->val!=q->val))     //利用||的短路效应避免运行时错误
            return false;
        return true;
    }
};

Solution 2 :非递归，待续