112 Path Sum

题目链接：https://leetcode.com/problems/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.

解题思路：

这道题的考点就是前序遍历二叉树。

1. 每遇到一个结点就用 sum 减去该结点的值，把剩余的值传递到当前结点的子树中。

2. 直到遇到叶子结点，判断此时 sum 是否已被减为 0，若被减为 0，说明存在该路径，否则说明当前遍历到的这条路径不符合要求。

要注意判断根结点为空的情况！

代码实现：

/**
* 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;
return helper(root, sum);
}
boolean helper(TreeNode root, int sum) {
if(root.left == null && root.right == null) {
if(sum - root.val == 0)
return true;
else
return false;
}
boolean left = false, right = false;
if(root.left != null)
left = helper(root.left, sum - root.val);
if(root.right != null)
right = helper(root.right, sum - root.val);
if(!right && !left)
return false;
else
return true;
}
}

114 / 114 test cases passed.
Status: Accepted
Runtime: 1 ms