[LeetCode] 113. Path Sum II

Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references.

A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: [[5,4,11,2],[5,8,4,5]]
Explanation: There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22

Example 2:

Input: root = [1,2,3], targetSum = 5
Output: []

Example 3:
Input: root = [1,2], targetSum = 0
Output: []

Constraints:
The number of nodes in the tree is in the range [0, 5000].
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000

路径总和 II。

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。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/path-sum-ii
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思路 - DFS

这题我只会用 DFS 做。这个题也有那么点 backtracking 的味道在里面，因为要求你输出所有的情况，需要回溯。同类型题还有 1457。

复杂度

时间O(n)
空间O(h)

代码

Java实现

class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        helper(res, new ArrayList<>(), root, sum);
        return res;
    }

    private void helper(List<List<Integer>> res, List<Integer> list, TreeNode root, int sum) {
        if (root == null) {
            return;
        }
        list.add(root.val);
        if (root.left == null && root.right == null) {
            if (sum == root.val) {
                res.add(new ArrayList<>(list));
            }
        }
        helper(res, list, root.left, sum - root.val);
        helper(res, list, root.right, sum - root.val);
        list.remove(list.size() - 1);
    }
}

JavaScript实现

/**
 * @param {TreeNode} root
 * @param {number} sum
 * @return {number[][]}
 */
var pathSum = function(root, sum) {
    let res = [];
    if (root === null) return res;
    helper(res, [], root, sum);
    return res;
};

var helper = function(res, list, root, sum) {
    if (root) {
        list.push(root.val);
        if (!root.left && !root.right && sum - root.val === 0) {
            res.push([...list]);
        }
        helper(res, list, root.left, sum - root.val);
        helper(res, list, root.right, sum - root.val);
        list.pop();
    }
    return res;
};

相关题目

112. Path Sum
113. Path Sum II
437. Path Sum III
1457. Pseudo-Palindromic Paths in a Binary Tree