112. Path Sum

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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);        }    }}

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