🌵113. Path Sum II
https://leetcode.com/problems/path-sum-ii/
js
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @param {number} sum
* @return {number[][]}
*/
function pathSum(root, sum) {
const result = []
calcSum(root, sum, [], result)
return result
}
function calcSum(root, sum, nums, result) {
if (!root) {
return
}
if (root.val === sum && !root.left && !root.right) {
result.push([...nums, root.val])
}
if (root.left) {
calcSum(root.left, sum - root.val, nums.concat([root.val]), result)
}
if (root.right) {
calcSum(root.right, sum - root.val, nums.concat([root.val]), result)
}
}py
# 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 pathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: List[List[int]]
"""
result = []
self.calcSum(root, sum, [], result)
return result
def calcSum(self, root, sum, nums, result):
if not root:
return
if root.val == sum and not root.left and not root.right:
result.append(nums + [root.val])
if root.left:
self.calcSum(root.left, sum - root.val, nums + [root.val], result)
if root.right:
self.calcSum(root.right, sum - root.val, nums + [root.val], result)go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func pathSum(root *TreeNode, sum int) [][]int {
result := [][]int{}
calcSum(root, sum, []int{}, &result)
return result
}
func calcSum(root *TreeNode, sum int, nums []int, result *[][]int) {
if root == nil {
return
}
nums = append(nums, root.Val)
if root.Val == sum && root.Left == nil && root.Right == nil {
*result = append(*result, nums)
}
if root.Left != nil {
calcSum(root.Left, sum-root.Val, append([]int{}, nums...), result)
}
if root.Right != nil {
calcSum(root.Right, sum-root.Val, append([]int{}, nums...), result)
}
}