112. Path Sum(判断路径和是否等于一个数)
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.
Note: A leaf is a node with no children.
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.
路径和定义为从 root 到 leaf 的所有节点的和。
方法一:递归
时间复杂度:o(n) 空间复杂度:O(1)
/** * 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&&sum==root.val) return true; return hasPathSum(root.left,sum-root.val)||hasPathSum(root.right,sum-root.val); } }
苟有恒,何必三更眠五更起;最无益,莫过一日暴十日寒。