112. 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

分情况讨论，如果一个node的left和right都为空，则此节点为leaf，通过判断节点数是否等于sum来返回true或false。如果一个节点为空，因为它的父节点不是leaf，则返回false。通过分治法，即可得

/** * 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) {            return root.val == sum;        }        boolean left = hasPathSum(root.left, sum - root.val);        boolean right = hasPathSum(root.right, sum - root.val);        return left || right;    }}