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.

Note: A leaf is a node with no children.

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.

路径和定义为从 root 到 leaf 的所有节点的和。

方法一：递归

时间复杂度：o（n）        空间复杂度：O(1)

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