add 116 solution

This commit is contained in:
3inchtime
2021-08-05 23:34:31 +08:00
parent 02999e422e
commit 7b7a39291a
3 changed files with 275 additions and 0 deletions

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package leetcode
type Node struct {
Val int
Left *Node
Right *Node
Next *Node
}
//解法一:迭代
func connect(root *Node) *Node {
if root == nil {
return root
}
q := []*Node{root}
for len(q) > 0 {
var p []*Node
// 遍历这一层的所有节点
for i, node := range q {
if i+1 < len(q) {
node.Next = q[i+1]
}
if node.Left != nil {
p = append(p, node.Left)
}
if node.Right != nil {
p = append(p, node.Right)
}
}
q = p
}
return root
}
// 解法二 递归
func connect2(root *Node) *Node {
if root == nil {
return nil
}
connectTwoNode(root.Left, root.Right)
return root
}
func connectTwoNode(node1, node2 *Node) {
if node1 == nil || node2 == nil {
return
}
node1.Next = node2
connectTwoNode(node1.Left, node1.Right)
connectTwoNode(node2.Left, node2.Right)
connectTwoNode(node1.Right, node2.Left)
}

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package leetcode
import (
"fmt"
"testing"
)
type question116 struct {
para116
ans116
}
// para 是参数
// one 代表第一个参数
type para116 struct {
one *Node
}
// ans 是答案
// one 代表第一个答案
type ans116 struct {
one *Node
}
func newQuestionNode()*Node{
node7 := &Node{}
node7.Val = 7
node6 := &Node{}
node6.Val = 6
node5 := &Node{}
node5.Val = 5
node4 := &Node{}
node4.Val = 4
node3 := &Node{}
node3.Val = 3
node2 := &Node{}
node2.Val = 2
node1 := &Node{}
node1.Val = 1
node1.Left = node2
node1.Right = node3
node2.Left = node4
node2.Right = node5
node3.Left = node6
node3.Right = node7
return node1
}
func newResultNode()*Node{
node7 := &Node{}
node7.Val = 7
node6 := &Node{}
node6.Val = 6
node5 := &Node{}
node5.Val = 5
node4 := &Node{}
node4.Val = 4
node3 := &Node{}
node3.Val = 3
node2 := &Node{}
node2.Val = 2
node1 := &Node{}
node1.Val = 1
node1.Left = node2
node1.Right = node3
node2.Left = node4
node2.Right = node5
node3.Left = node6
node3.Right = node7
node1.Next = nil
node2.Next = node3
node3.Next = nil
node4.Next = node5
node5.Next = node6
node6.Next = node7
node7.Next = nil
return node1
}
func Test_Problem116(t *testing.T) {
qs := []question116{
{
para116{newQuestionNode()},
ans116{newResultNode()},
},
}
fmt.Printf("------------------------Leetcode Problem 116------------------------\n")
for _, q := range qs {
_, p := q.ans116, q.para116
fmt.Printf("【input】:%v ", p.one)
fmt.Printf("【output】:%v \n", connect(p.one))
}
fmt.Printf("\n\n\n")
}

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# [116. Distinct Subsequences](https://leetcode.com/problems/populating-next-right-pointers-in-each-node/)
## 题目
You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:
```go
struct Node {
int val;
Node *left;
Node *right;
Node *next;
}
```
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
Initially, all next pointers are set to NULL.
**Follow up:**
- You may only use constant extra space.
- Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.
**Example 1:**
```
Input: root = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
```
**Constraints:**
- The number of nodes in the given tree is less than 4096.
- -1000 <= node.val <= 1000
## 题目大意
将二叉树的每一层节点都连接起来形成一个链表每层的最后一个节点指向NULL.
## 解题思路
本质上是二叉树的层序遍历,基于广度优先搜索,将每层的节点放入队列,并遍历队列进行连接。
## 代码
```go
package leetcode
// 解法一:迭代
func connect(root *Node) *Node {
if root == nil {
return root
}
q := []*Node{root}
for len(q) > 0 {
var p []*Node
for i, node := range q {
if i+1 < len(q) {
node.Next = q[i+1]
}
if node.Left != nil {
p = append(p, node.Left)
}
if node.Right != nil {
p = append(p, node.Right)
}
}
q = p
}
return root
}
// 解法二 递归
func connect2(root *Node) *Node {
if root == nil {
return nil
}
connectTwoNode(root.Left, root.Right)
return root
}
func connectTwoNode(node1, node2 *Node) {
if node1 == nil || node2 == nil {
return
}
node1.Next = node2
connectTwoNode(node1.Left, node1.Right)
connectTwoNode(node2.Left, node2.Right)
connectTwoNode(node1.Right, node2.Left)
}
```