題目：

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1    、
   / \   
  2   2  
 / \ / \ 
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

  1    
 / \   
2   2    
 \   \    
 3    3
 

Note: Bonus points if you could solve it both recursively and iteratively.

解釋： 判斷一顆二叉樹是否是映象的，用dfs。不一定非要再另外寫一個函式，直接在本來的函式裡面寫遞迴即可。 python程式碼：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
 


class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        def helper(p,q):
            if p==None and q==None:
                return True
            elif p==None or q==None:
                return False
            return
 
 p.val ==q.val and helper(p.left,q.right) and helper(p.right,q.left)     
        if root==None:
            return True
        return helper(root.left,root.right)

c++程式碼：

/**
 * 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 helper(root->left,root->right);
    }
    bool helper(TreeNode* p,TreeNode* q)
    {
        if (p==NULL&& q==NULL)
            return true;
        else if(p==NULL ||q==NULL)
            return false;
        return (p->val==q->val) &&helper(p->left,q->right)&&helper(p->right,q->left);
    }
};

總結： 反正樹的題目遞迴就對啦，對於特殊情況需要好好考慮一下。