34. 二叉树中和为某一值的路径

comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/lcof/%E9%9D%A2%E8%AF%95%E9%A2%9834.%20%E4%BA%8C%E5%8F%89%E6%A0%91%E4%B8%AD%E5%92%8C%E4%B8%BA%E6%9F%90%E4%B8%80%E5%80%BC%E7%9A%84%E8%B7%AF%E5%BE%84/README.md

面试题 34. 二叉树中和为某一值的路径

题目描述

给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

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,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]

示例 2：

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

示例 3：

输入：root = [1,2], targetSum = 0
输出：[]

提示：

树中节点总数在范围 [0, 5000] 内
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000

注意：本题与主站 113 题相同：https://leetcode.cn/problems/path-sum-ii/

解法

方法一：递归

从根节点开始，递归遍历每个节点，每次递归时，将当前节点值加入到路径中，然后判断当前节点是否为叶子节点，如果是叶子节点并且路径和等于目标值，则将该路径加入到结果中。如果当前节点不是叶子节点，则递归遍历其左右子节点。递归遍历时，需要将当前节点从路径中移除，以确保返回父节点时路径刚好是从根节点到父节点。

时间复杂度 $O(n^2)$ ，空间复杂度 $O (n)$ 。其中 $n$ 是二叉树的节点数。

Python3

# 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 pathSum(self, root: TreeNode, target: int) -> List[List[int]]:
        def dfs(root, s):
            if root is None:
                return
            
            #加入节点的同时，检查路径是否满足
            t.append(root.val)
            s -= root.val
            if root.left is None and root.right is None and s == 0:#root不能为空
                ans.append(t[:]) #t会动态改变
                
            dfs(root.left, s)
            dfs(root.right, s)
            t.pop() #这种写法，回溯时不需要s+=root.val

        ans = []
        t = []
        dfs(root, target)
        return ans

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

class Solution(object):
    def findPath(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        def dfs(root):
            nonlocal s
            if root is None:
                return
            
            t.append(root.val)
            s-=root.val
            if root.left is None and root.right is None and s==0:
                  ans.append(t[:])
            
            dfs(root.left)
            dfs(root.right)
            t.pop()
            s+=root.val

        ans = []
        t=[]
        s=sum
        dfs(root)
        return ans

Java

/**
 * 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 {
    private List<Integer> t = new ArrayList<>();
    private List<List<Integer>> ans = new ArrayList<>();

    public List<List<Integer>> pathSum(TreeNode root, int target) {
        dfs(root, target);
        return ans;
    }

    private void dfs(TreeNode root, int s) {
        if (root == null) {
            return;
        }
        t.add(root.val);
        s -= root.val;
        if (root.left == null && root.right == null && s == 0) {
            ans.add(new ArrayList<>(t));
        }
        dfs(root.left, s);
        dfs(root.right, s);
        t.remove(t.size() - 1);
    }
}

C++

/**
 * 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:
    vector<vector<int>> pathSum(TreeNode* root, int target) {
        vector<vector<int>> ans;
        vector<int> t;
        function<void(TreeNode * root, int s)> dfs = [&](TreeNode* root, int s) {
            if (!root) {
                return;
            }
            t.push_back(root->val);
            s -= root->val;
            if (!root->left && !root->right && !s) {
                ans.push_back(t);
            }
            dfs(root->left, s);
            dfs(root->right, s);
            t.pop_back();
        };
        dfs(root, target);
        return ans;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func pathSum(root *TreeNode, target int) (ans [][]int) {
	t := []int{}
	var dfs func(*TreeNode, int)
	dfs = func(root *TreeNode, s int) {
		if root == nil {
			return
		}
		t = append(t, root.Val)
		s -= root.Val
		if root.Left == nil && root.Right == nil && s == 0 {
			ans = append(ans, slices.Clone(t))
		}
		dfs(root.Left, s)
		dfs(root.Right, s)
		t = t[:len(t)-1]
	}
	dfs(root, target)
	return
}

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 pathSum(root: TreeNode | null, target: number): number[][] {
    const ans: number[][] = [];
    const t: number[] = [];

    const dfs = (root: TreeNode | null, s: number): void => {
        if (!root) {
            return;
        }
        const { val, left, right } = root;
        t.push(val);
        s -= val;
        if (!left && !right && s === 0) {
            ans.push([...t]);
        }
        dfs(left, s);
        dfs(right, s);
        t.pop();
    };

    dfs(root, target);
    return ans;
}

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::cell::RefCell;
use std::rc::Rc;
impl Solution {
    fn dfs(
        root: &Option<Rc<RefCell<TreeNode>>>,
        mut target: i32,
        t: &mut Vec<i32>,
        ans: &mut Vec<Vec<i32>>,
    ) {
        if let Some(node) = root.as_ref() {
            let node = node.borrow();
            t.push(node.val);
            target -= node.val;
            if node.left.is_none() && node.right.is_none() && target == 0 {
                ans.push(t.clone());
            }
            Self::dfs(&node.left, target, t, ans);
            Self::dfs(&node.right, target, t, ans);
            t.pop();
        }
    }

    pub fn path_sum(root: Option<Rc<RefCell<TreeNode>>>, target: i32) -> Vec<Vec<i32>> {
        let mut ans = vec![];
        Self::dfs(&root, target, &mut vec![], &mut ans);
        ans
    }
}

JavaScript

/**
 * 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} target
 * @return {number[][]}
 */
var pathSum = function (root, target) {
    const ans = [];
    const t = [];
    const dfs = (root, s) => {
        if (!root) {
            return;
        }
        const { val, left, right } = root;
        t.push(val);
        s -= val;
        if (!left && !right && !s) {
            ans.push([...t]);
        }
        dfs(left, s);
        dfs(right, s);
        t.pop();
    };
    dfs(root, target);
    return ans;
};

C#

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    private List<IList<int>> ans = new List<IList<int>>();
    private List<int> t = new List<int>();

    public IList<IList<int>> PathSum(TreeNode root, int target) {
        dfs(root, target);
        return ans;
    }

    private void dfs(TreeNode root, int s) {
        if (root == null) {
            return;
        }
        t.Add(root.val);
        s -= root.val;
        if (root.left == null && root.right == null && s == 0) {
            ans.Add(new List<int>(t));
        }
        dfs(root.left, s);
        dfs(root.right, s);
        t.RemoveAt(t.Count - 1);
    }
}

Swift

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init() { self.val = 0; self.left = nil; self.right = nil; }
 *     public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
 *     public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
 *         self.val = val
 *         self.left = left
 *         self.right = right
 *     }
 * }
 */

class Solution {
    private var t = [Int]()
    private var ans = [[Int]]()

    func pathSum(_ root: TreeNode?, _ target: Int) -> [[Int]] {
        dfs(root, target)
        return ans
    }

    private func dfs(_ root: TreeNode?, _ s: Int) {
        guard let root = root else {
            return
        }
        t.append(root.val)
        let remainingSum = s - root.val
        if root.left == nil && root.right == nil && remainingSum == 0 {
            ans.append(Array(t))
        }
        dfs(root.left, remainingSum)
        dfs(root.right, remainingSum)
        t.removeLast()
    }
}