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 。如果存在，返回 true ；否则，返回 false 。

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
解释：等于目标和的根节点到叶节点路径如上图所示。

示例 2：

输入：root = [1,2,3], targetSum = 5
输出：false
解释：树中存在两条根节点到叶子节点的路径：
(1 --> 2): 和为 3
(1 --> 3): 和为 4
不存在 sum = 5 的根节点到叶子节点的路径。

示例 3：

输入：root = [], targetSum = 0
输出：false
解释：由于树是空的，所以不存在根节点到叶子节点的路径。

 

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);
    }
}

C++

class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {

        if(root == NULL)return false;
        if(root->right == NULL && root->left == NULL && sum == root->val)return true;

        bool leftTrue = hasPathSum(root->left,sum - root->val);
        bool rightTrue = hasPathSum(root->right,sum - root->val);

        return (leftTrue || rightTrue);
    }
};

TypeScript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function hasPathSum(root: TreeNode | null, targetSum: number): boolean {
    if (root == null) {
        return false;
    }
    const { val, left, right } = root;
    if (left == null && right == null) {
        return targetSum - val === 0;
    }
    return (
        hasPathSum(left, targetSum - val) || hasPathSum(right, targetSum - val)
    );
}

Rust

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    pub fn has_path_sum(root: Option<Rc<RefCell<TreeNode>>>, target_sum: i32) -> bool {
        match root {
            None => false,
            Some(node) => {
                let mut node = node.borrow_mut();
                // 确定叶结点身份
                if node.left.is_none() && node.right.is_none() {
                    return target_sum - node.val == 0;
                }
                let val = node.val;
                Self::has_path_sum(node.left.take(), target_sum - val)
                    || Self::has_path_sum(node.right.take(), target_sum - val)
            }
        }
    }
}

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