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  • Symmetric Tree

    这里A这道题的时候,刚看到题目的时候想到用递归。后面题目也有提示,用递归或迭代。还是没有弄出来,主要参考了http://www.cnblogs.com/remlostime/archive/2012/11/15/2772230.html

    LeetCode Oj:https://oj.leetcode.com/problems/

    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}".
     
     
     1 class TreeNode {
     2      int val;
     3      TreeNode left;
     4      TreeNode right;
     5      TreeNode(int x) { val = x; }
     6   }
     7 
     8 public class Solution {
     9     public boolean isSymmetric(TreeNode root) {        
    10         if(null == root)
    11             return true;
    12         return check(root.left, root.right);
    13     }
    14     public boolean check(TreeNode leftNode, TreeNode rightNode){//开始有想到这样,用左右子节点
    15         if(null == leftNode && null == rightNode) //两个都是空节点
    16             return true;
    17         if(null == leftNode || null == rightNode)//其中一个空节点
    18             return false;
    19         //两个都不是空节点
    20         return leftNode.val == rightNode.val && check(leftNode.left, rightNode.right) 
    21                 && check(leftNode.right, rightNode.left);//这里用的很漂亮,主要逻辑也在这里
    22     }
    23 }
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  • 原文地址:https://www.cnblogs.com/luckygxf/p/4049976.html
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