112. Path Sum
112. Path Sum
🟢 🔖 树
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二叉树
🔗 LeetCode
题目
Given the root
of a binary tree and an integer targetSum
, return true
if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum
.
A leaf is a node with no children.
Example 1:
![](https://assets.leetcode.com/uploads/2021/01/18/pathsum1.jpg)
Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation: The root-to-leaf path with the target sum is shown.
Example 2:
![](https://assets.leetcode.com/uploads/2021/01/18/pathsum2.jpg)
Input: root = [1,2,3], targetSum = 5
Output: false
Explanation: There two root-to-leaf paths in the tree:
(1 --> 2): The sum is 3.
(1 --> 3): The sum is 4.
There is no root-to-leaf path with sum = 5.
Example 3:
Input: root = [], targetSum = 0
Output: false
Explanation: Since the tree is empty, there are no root-to-leaf paths.
Constraints:
- The number of nodes in the tree is in the range
[0, 5000]
. -1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
题目大意
给定一个二叉树和一个目标和,判断该树中是否存在根节点到叶子节点的路径,这条路径上所有节点值相加等于目标和。说明: 叶子节点是指没有子节点的节点。
解题思路
递归求解即可
代码
/**
* @param {TreeNode} root
* @param {number} targetSum
* @return {boolean}
*/
var hasPathSum = function (root, targetSum) {
if (!root) return false;
if (!root.left && !root.right) return root.val == targetSum;
return (
hasPathSum(root.left, targetSum - root.val) ||
hasPathSum(root.right, targetSum - root.val)
);
};
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- [129. 求根节点到叶节点数字之和](./0129.md)
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