112. Path Sum

题目：

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.

思路：

直觉的思路就是用递归。。。写出了下述待优化的代码。。。

代码：

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if(root == null){return false;}
sum -= root.val;
if(root.left == null && root.right == null){
if(sum == 0){
return true;
}else{
return false;
}
}else{
boolean left = false, right = false;
if(root.left != null){left = hasPathSum(root.left, sum);}
if(root.right != null){right = hasPathSum(root.right, sum);}
return (left||right);
}
}
}