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文章目录
  1. 1. Symmetric Tree
    1. 1.1. 题目
    2. 1.2. 思路
    3. 1.3. 解题

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.

思路

逐个比较两个对称结点

解题

c++版

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

python版

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

class Solution:
    # @param root, a tree node
    # @return a boolean
    def isSymmetric(self, root):
        if not root:
            return True
        else:
            return self.traverse(root.left,root.right)
    
    def traverse(self,left,right):
        if not left and not right:
            return True
        elif not left and right:
            return False
        elif not right and left:
            return False
        elif right.val != left.val:
            return False
        else:
            return self.traverse(left.left,right.right) and self.traverse(left.right,right.left)
leetcode
leetcode,tree
文章目录
  1. 1. Symmetric Tree
    1. 1.1. 题目
    2. 1.2. 思路
    3. 1.3. 解题