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257.二叉树的所有路径.go
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257.二叉树的所有路径.go
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import (
"strconv"
"strings"
)
/*
* @lc app=leetcode.cn id=257 lang=golang
*
* [257] 二叉树的所有路径
*
* https://leetcode-cn.com/problems/binary-tree-paths/description/
*
* algorithms
* Easy (58.25%)
* Likes: 123
* Dislikes: 0
* Total Accepted: 9.4K
* Total Submissions: 16.2K
* Testcase Example: '[1,2,3,null,5]'
*
* 给定一个二叉树,返回所有从根节点到叶子节点的路径。
*
* 说明: 叶子节点是指没有子节点的节点。
*
* 示例:
*
* 输入:
*
* 1
* / \
* 2 3
* \
* 5
*
* 输出: ["1->2->5", "1->3"]
*
* 解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3
*
*/
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func binaryTreePaths(root *TreeNode) []string {
paths := binaryPath(root)
result := make([]string, 0)
for _, path := range paths {
result = append(result, strings.Join(path, "->"))
}
return result
}
func binaryPath(node *TreeNode) [][]string {
if node == nil {
return [][]string{}
}
if node.Left == nil && node.Right == nil {
return [][]string{
[]string{strconv.Itoa(node.Val)},
}
}
if node.Left == nil && node.Right != nil {
result := make([][]string, 0)
rightResult := binaryPath(node.Right)
for _, right := range rightResult {
result = append(result, append([]string{strconv.Itoa(node.Val)}, right...))
}
return result
}
if node.Left != nil && node.Right == nil {
result := make([][]string, 0)
leftResult := binaryPath(node.Left)
for _, left := range leftResult {
result = append(result, append([]string{strconv.Itoa(node.Val)}, left...))
}
return result
} else {
result := make([][]string, 0)
leftResult := binaryPath(node.Left)
for _, left := range leftResult {
result = append(result, append([]string{strconv.Itoa(node.Val)}, left...))
}
rightResult := binaryPath(node.Right)
for _, right := range rightResult {
result = append(result, append([]string{strconv.Itoa(node.Val)}, right...))
}
return result
}
}