Symmetric Tree 深度优先搜索

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}".

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Tree Depth-first Search

 递归，保存左右两个节点，然后判断leftNode->left和rightNode->right，以及leftNode->right和rightNode->left。如此不断递归

/**
 * 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 dfscheck(TreeNode *leftNode,TreeNode *rightNode){
        if(leftNode==NULL &&rightNode==NULL)
            return true;
        if(leftNode==NULL || rightNode==NULL)
            return false;
        return leftNode->val==rightNode->val && dfscheck(leftNode->left,rightNode->right) && dfscheck(leftNode->right,rightNode->left);
    }
    bool isSymmetric(TreeNode *root) {
        if(root==NULL)
            return true;
        return dfscheck(root->left,root->right);
    }
};