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.

 1 /**
 2  * Definition for binary tree
 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     boolean has;
12     public boolean hasPathSum(TreeNode root, int sum) {
13         // Start typing your Java solution below
14         // DO NOT write main() function
15         has = false;
16         path(root, sum);
17         return has;
18     }
19     public void path(TreeNode r, int s){
20         if(r == null) return;
21         if(r.left == null && r.right == null){
22             if(r.val == s)has = true;
23                 return;
24         }
25         else{
26             path(r.right, s - r.val);
27             path(r.left, s - r.val);
28         }
29     }
30 }

第二遍：

 1 public class Solution {
 2     public boolean hasPathSum(TreeNode root, int sum) {
 3         // Note: The Solution object is instantiated only once and is reused by each test case.
 4         if(root == null) return false;
 5         sum -= root.val;
 6         if(root.left == null && root.right == null){
 7             if(sum == 0) return true;
 8             else return false;
 9         }
10         return hasPathSum(root.left , sum) || hasPathSum(root.right, sum);
11     }
12 }

if(root == null) return false; 很重要！