Home Original page

leetcode/solution/0100-0199/0112.Path Sum/README.md at main · rutonggaoya/leetcode

112. 路径总和

English Version

题目描述

给你二叉树的根节点 root 和一个表示目标和的整数 targetSum ，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 targetSum 。

叶子节点 是指没有子节点的节点。

示例 1：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
输出：true

示例 2：

输入：root = [1,2,3], targetSum = 5
输出：false

示例 3：

输入：root = [1,2], targetSum = 0
输出：false

提示：

  • 树中节点的数目在范围 [0, 5000] 内
  • -1000 <= Node.val <= 1000
  • -1000 <= targetSum <= 1000

解法

递归求解，递归地询问它的子节点是否能满足条件即可。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        def dfs(root, sum):
            if root is None:
                return False
            if root.val == sum and root.left is None and root.right is None:
                return True
            return dfs(root.left, sum - root.val) or dfs(root.right, sum - root.val)
        return dfs(root, sum)

Java

/**
 * 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) {
        return dfs(root, sum);
    }

    private boolean dfs(TreeNode root, int sum) {
        if (root == null) return false;
        if (root.val == sum && root.left == null && root.right == null) return true;
        return dfs(root.left, sum - root.val) || dfs(root.right, sum - root.val);
    }
}

...