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.
给定一棵树和一个数,判断这个数是不是从根节点到叶子节点的和。
递归实现很简单。
/** * 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.left == null && root.right == null) return sum==root.val; return hasPathSum(root.left,sum-root.val)||hasPathSum(root.right,sum-root.val); } }