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.

 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         boolean flag=false;
13         
14         if(root==null)
15         return false;
16         
17         if(root.left==null&&root.right==null&&root.val==sum)
18         {
19         flag=true;
20         return flag;
21         }
22         
23         if(root.left!=null||root.right!=null)
24         {
25             if(root.left!=null&&flag==false)
26             {
27              flag=hasPathSum(root.left,sum-root.val);   
28              if(flag==true)
29              return flag;
30             }
31             
32             if(root.right!=null&&flag==false)
33             {
34              flag=hasPathSum(root.right,sum-root.val);   
35              if(flag==true)
36              return flag;
37             }
38             
39         }
40         return flag;
41     }
42 }

            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：

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