LeetCode OJ 112. Path Sum

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.

代码如下：

 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     public boolean hasPathSum(TreeNode root, int sum) {
12         if(root == null) return false;
13         if(root.left==null && root.right==null){
14             if(root.val==sum) return true;
15             else return false;
16         }
17         
18         boolean left = false;
19         boolean right = false;
20         if(root.left != null){
21             left = hasPathSum(root.left, sum - root.val);
22         }
23         if(root.right != null){
24             right = hasPathSum(root.right, sum - root.val);
25         }
26         
27         if(left==true || right==true) return true;
28         else return false;
29     }
30 }

 

posted @ 2016-04-28 10:32  Black_Knight  阅读(177)  评论(0)  编辑  收藏  举报