[LeetCode] 112. Path Sum
2022/1/9 6:07:04
本文主要是介绍[LeetCode] 112. Path Sum,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
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.
Example1:
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.
Example2:
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.
Example3:
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
这道题的思路一看就是递归,但是需要注意的递归的出口,不但要处理null的情况,更要处理叶子结点。targetSum每次都会减掉当时的层结点的值,当到叶子的时候,只要叶子的值和剩下的targetSum一致,就可以了。其实就是前序遍历,每次把当前结点的val减掉。
注意:
- 这道题必须是到叶子结点累加,所以不需要考虑如果中间过程中有累加与targetSum相等。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public boolean hasPathSum(TreeNode root, int targetSum) { if (root == null) { return false; } targetSum -= root.val; if (root.left == null && root.right == null) { return targetSum == 0; } return hasPathSum(root.left, targetSum) || hasPathSum(root.right, targetSum); } }
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