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！