112. 路径总和

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

方法一：递归

从根节点开始，递归地对树进行遍历，并在遍历过程中更新节点的值为从根节点到该节点的路径和。当遍历到叶子节点时，判断该路径和是否等于目标值，如果相等则返回 true，否则返回 false。

时间复杂度 $O(n)$，其中 $n$ 是二叉树的节点数。对每个节点访问一次。

Python3JavaC++GoTypeScriptRustJavaScript

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
        def dfs(root, s):
            if root is None:
                return False
            s += root.val
            if root.left is None and root.right is None and s == targetSum:
                return True
            return dfs(root.left, s) or dfs(root.right, s)

        return dfs(root, 0)

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

    private boolean dfs(TreeNode root, int s) {
        if (root == null) {
            return false;
        }
        s -= root.val;
        if (root.left == null && root.right == null && s == 0) {
            return true;
        }
        return dfs(root.left, s) || dfs(root.right, s);
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode* root, int targetSum) {
        function<bool(TreeNode*, int)> dfs = [&](TreeNode* root, int s) -> int {
            if (!root) return false;
            s += root->val;
            if (!root->left && !root->right && s == targetSum) return true;
            return dfs(root->left, s) || dfs(root->right, s);
        };
        return dfs(root, 0);
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func hasPathSum(root *TreeNode, targetSum int) bool {
    var dfs func(*TreeNode, int) bool
    dfs = func(root *TreeNode, s int) bool {
        if root == nil {
            return false
        }
        s += root.Val
        if root.Left == nil && root.Right == nil && s == targetSum {
            return true
        }
        return dfs(root.Left, s) || dfs(root.Right, s)
    }
    return dfs(root, 0)
}

/**
 * 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);
}

// 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::cell::RefCell;
use std::rc::Rc;
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)
            }
        }
    }
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {boolean}
 */
var hasPathSum = function (root, targetSum) {
    function dfs(root, s) {
        if (!root) return false;
        s += root.val;
        if (!root.left && !root.right && s == targetSum) return true;
        return dfs(root.left, s) || dfs(root.right, s);
    }
    return dfs(root, 0);
};

112. 路径总和

题目描述

解法

方法一：递归

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