Path Sum

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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.

描述

判断是否有一条路径（从根节点到叶子节点）的和为 sum

分析

类似于树的前序遍历

代码

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/**
 * 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;
        }

        if (root.val == sum && root.left == null && root.right == null) {
            return true;
        }

        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}