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.
time: O(n), space: O(height)
/** * 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) { return sum - root.val == 0; } return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val); } }
原文：https://www.cnblogs.com/fatttcat/p/10198680.html