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LeetCode 112. Path Sum (二叉树路径之和)

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,
              5
             /             4   8
           /   /           11  13  4
         /  \              7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

 


题目标签:Tree

  这道题目给了我们一个二叉树和一个sum, 让我们判断这个二叉树是否有至少一条path 的之和是等于sum的。利用preOrder 来遍历树,每次用sum 减去当前点的值,每当遇到一个leaf node 的时候检查sum 是不是等于0, 返回ture 和false。利用 || 来return 所有的boolean 值, 至少有过一个true,一个path之和等于sum, 总的boolean 就是true。

 

Java Solution:

Runtime beats 13.93% 

完成日期:07/03/2017

关键词:Tree

关键点:当是leaf node 的时候检查sum;利用 || return两个children的返回值

 

 

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution 
11 {
12     public boolean hasPathSum(TreeNode root, int sum) 
13     {
14         if(root == null)
15             return false;
16         
17         sum -= root.val;
18         
19         if(root.left == null && root.right == null)
20         {
21             if(sum == 0)
22                 return true;
23             else 
24                 return false;
25         }
26         
27         return hasPathSum(root.left, sum) || hasPathSum(root.right, sum);
28     }
29 }

参考资料:

http://www.cnblogs.com/springfor/p/3879825.html

LeetCode 112. Path Sum (二叉树路径之和)