判断一颗二叉树中是否存在一天路径(从根节点到叶子节点)的val值之和等于给定的sum。注意判断root为null的时候。
成也递归败也递归。。。。。
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public boolean hasPathSum(TreeNode root, int sum) { return PathSum(root, 0, sum); } public boolean PathSum(TreeNode root, int val, int sum) { if(root == null) return false; val += root.val; if(root.left==null && root.right==null) { if(val == sum) { return true; } else { return false; } } return PathSum(root.left, val, sum) || PathSum(root.right, val, sum); } }