113. Path Sum II

Medium

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

解題

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func pathSum(root *TreeNode, targetSum int) [][]int {
    ans := make([][]int, 0)

    var dfs func(*TreeNode, int, []int)
    dfs = func(n *TreeNode, target int, cur []int) {
        if n == nil {
            return 
        }

        if target == n.Val && n.Left == nil && n.Right == nil {
            newarr := append([]int{}, cur...)
            ans = append(ans, append(newarr, n.Val))
            return
        }

        dfs(n.Left, target - n.Val, append(cur, n.Val))
        dfs(n.Right, target - n.Val, append(cur, n.Val))
    }

    dfs(root, targetSum, []int{})

    return ans
}