Add solution 0589

leetcode/0589.N-ary-Tree-Preorder-Traversal/589. N-ary Tree Preorder Traversal.go

@@ -0,0 +1,43 @@
+package leetcode
+
+//  Definition for a Node.
+type Node struct {
+	Val      int
+	Children []*Node
+}
+
+// 解法一 非递归
+func preorder(root *Node) []int {
+	res := []int{}
+	if root == nil {
+		return res
+	}
+	stack := []*Node{root}
+	for len(stack) > 0 {
+		r := stack[len(stack)-1]
+		stack = stack[:len(stack)-1]
+		res = append(res, r.Val)
+		tmp := []*Node{}
+		for _, v := range r.Children {
+			tmp = append([]*Node{v}, tmp...) // 逆序存点
+		}
+		stack = append(stack, tmp...)
+	}
+	return res
+}
+
+// 解法二 递归
+func preorder1(root *Node) []int {
+	res := []int{}
+	preorderdfs(root, &res)
+	return res
+}
+
+func preorderdfs(root *Node, res *[]int) {
+	if root != nil {
+		*res = append(*res, root.Val)
+		for i := 0; i < len(root.Children); i++ {
+			preorderdfs(root.Children[i], res)
+		}
+	}
+}

leetcode/0589.N-ary-Tree-Preorder-Traversal/589. N-ary Tree Preorder Traversal_test.go

@@ -0,0 +1,82 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+type question589 struct {
+	para589
+	ans589
+}
+
+// para 是参数
+// one 代表第一个参数
+type para589 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans589 struct {
+	one []int
+}
+
+func Test_Problem589(t *testing.T) {
+
+	qs := []question589{
+
+		{
+			para589{[]int{1, structures.NULL, 3, 2, 4, structures.NULL, 5, 6}},
+			ans589{[]int{1, 3, 5, 6, 2, 4}},
+		},
+
+		{
+			para589{[]int{1, structures.NULL, 2, 3, 4, 5, structures.NULL, structures.NULL, 6, 7, structures.NULL, 8, structures.NULL, 9, 10, structures.NULL, structures.NULL, 11, structures.NULL, 12, structures.NULL, 13, structures.NULL, structures.NULL, 14}},
+			ans589{[]int{1, 2, 3, 6, 7, 11, 14, 4, 8, 12, 5, 9, 13, 10}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 589------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans589, q.para589
+		fmt.Printf("【input】:%v      ", p)
+		rootOne := int2NaryNode(p.one)
+		fmt.Printf("【output】:%v      \n", preorder(rootOne))
+	}
+	fmt.Printf("\n\n\n")
+}
+
+func int2NaryNode(nodes []int) *Node {
+	root := &Node{}
+	if len(nodes) > 1 {
+		root.Val = nodes[0]
+	}
+	queue := []*Node{}
+	queue = append(queue, root)
+	i := 1
+	count := 0
+	for i < len(nodes) {
+		node := queue[0]
+
+		childrens := []*Node{}
+		for ; i < len(nodes) && nodes[i] != structures.NULL; i++ {
+			tmp := &Node{Val: nodes[i]}
+			childrens = append(childrens, tmp)
+			queue = append(queue, tmp)
+		}
+		count++
+		if count%2 == 0 {
+			queue = queue[1:]
+			count = 1
+		}
+		if node != nil {
+			node.Children = childrens
+		}
+		i++
+	}
+	return root
+}

leetcode/0589.N-ary-Tree-Preorder-Traversal/README.md

@@ -0,0 +1,90 @@
+# [589. N-ary Tree Preorder Traversal](https://leetcode.com/problems/n-ary-tree-preorder-traversal/)
+
+## 题目
+
+Given the `root` of an n-ary tree, return *the preorder traversal of its nodes' values*.
+
+Nary-Tree input serialization is represented in their level order traversal. Each group of children is separated by the null value (See examples)
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png](https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png)
+
+```
+Input: root = [1,null,3,2,4,null,5,6]
+Output: [1,3,5,6,2,4]
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png](https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png)
+
+```
+Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
+Output: [1,2,3,6,7,11,14,4,8,12,5,9,13,10]
+```
+
+**Constraints:**
+
+- The number of nodes in the tree is in the range `[0, 104]`.
+- `0 <= Node.val <= 10^4`
+- The height of the n-ary tree is less than or equal to `1000`.
+
+**Follow up:** Recursive solution is trivial, could you do it iteratively?
+
+## 题目大意
+
+给定一个 N 叉树，返回其节点值的 **前序遍历** 。N 叉树 在输入中按层序遍历进行序列化表示，每组子节点由空值 `null` 分隔（请参见示例）。
+
+## 解题思路
+
+- N 叉树和二叉树的前序遍历原理完全一样。二叉树非递归解法需要用到栈辅助，N 叉树同样如此。将父节点的所有孩子节点**逆序**入栈，逆序的目的是为了让前序节点永远在栈顶。依次循环直到栈里所有元素都出栈。输出的结果即为 N 叉树的前序遍历。时间复杂度 O(n)，空间复杂度 O(n)。
+- 递归解法非常简单，见解法二。
+
+## 代码
+
+```go
+package leetcode
+
+//  Definition for a Node.
+type Node struct {
+	Val      int
+	Children []*Node
+}
+
+// 解法一 非递归
+func preorder(root *Node) []int {
+	res := []int{}
+	if root == nil {
+		return res
+	}
+	stack := []*Node{root}
+	for len(stack) > 0 {
+		r := stack[len(stack)-1]
+		stack = stack[:len(stack)-1]
+		res = append(res, r.Val)
+		tmp := []*Node{}
+		for _, v := range r.Children {
+			tmp = append([]*Node{v}, tmp...) // 逆序存点
+		}
+		stack = append(stack, tmp...)
+	}
+	return res
+}
+
+// 解法二 递归
+func preorder1(root *Node) []int {
+	res := []int{}
+	preorderdfs(root, &res)
+	return res
+}
+
+func preorderdfs(root *Node, res *[]int) {
+	if root != nil {
+		*res = append(*res, root.Val)
+		for i := 0; i < len(root.Children); i++ {
+			preorderdfs(root.Children[i], res)
+		}
+	}
+}
+```