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.
中文(判断二叉树一条路径和是否等于给定的值)
代码:
/** * Definition for a binary tree node. * public class TreeNode { * public int val; * public TreeNode left; * public TreeNode right; * public TreeNode(int x) { val = x; } * } */ public class Solution { public bool HasPathSum(TreeNode root, int sum) { if(root == null) return false; if(root.left == null && root.right == null && root.val == sum) return true; else if(root.left == null && root.right == null && root.val != sum) return false; return HasPathSum(root.left, sum - root.val) || HasPathSum(root.right, sum - root.val); } }