113.Path Sum II
Loading...leetcode.com1、 先读题，题目是求从根到叶子node，路径上所有数字之和等于sum 的所有路径。
2、先求出从root 到叶子node，所有路径的二维数组，再判断二维数组中那些元素之和等于sum值。
3、用递归深度优先搜索解决，用一个二维数组统计所有路径，一个一维数组保存每一条路径。
4、当到达叶子节点时，将一维数组保存的路径添加到二维数组中。
请看黄哥所写的Python 代码和Go 语言代码。
Python 代码
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# 黄哥Python培训 黄哥所写
class Solution:
def pathSum(self, root: TreeNode, s: int) -> List[List[int]]:
if root is None:
return []
paths = []
def dfs(root, nodes):
if root is None:
return
nodes.append(root.val)
if root.left is None and root.right is None:
paths.append(nodes)
dfs(root.left, nodes[:])
dfs(root.right, nodes[:])
dfs(root, [])
return [item for item in paths if sum(item) == s]Go 代码
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
// 黄哥Python培训 黄哥所写
func pathSum(root *TreeNode, sum int) [][]int {
if root == nil {
return [][]int{}
}
var paths, res [][]int
path := []int{}
dfs(root, path, &paths)
for _, item := range paths {
if sumSlice(item) == sum {
res = append(res, item)
}
}
return res
}
func dfs(root *TreeNode, path []int, paths *[][]int) {
if root == nil {
return
}
path = append(path, root.Val)
if root.Left == nil && root.Right == nil {
copy := append(make([]int, 0, len(path)), path...)
*paths = append(*paths, copy)
}
dfs(root.Left, path, paths)
dfs(root.Right, path, paths)
}
func sumSlice(arr []int) int{
res := 0
for _, item := range arr {
res += item
}
return res
}
