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Commit 7bbc8ba

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Merge pull request #406 from X-shuffle/master
添加0071.0501.二叉搜索树中的众数 go版本
2 parents e332190 + 2f514cd commit 7bbc8ba

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‎problems/0235.二叉搜索树的最近公共祖先.md‎

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@@ -265,6 +265,54 @@ class Solution:
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else: return root
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```
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Go:
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> BSL法
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```go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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//利用BSL的性质(前序遍历有序)
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func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
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if root==nil{return nil}
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if root.Val>p.Val&&root.Val>q.Val{//当前节点的值大于给定的值,则说明满足条件的在左边
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return lowestCommonAncestor(root.Left,p,q)
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}else if root.Val<p.Val&&root.Val<q.Val{//当前节点的值小于各点的值,则说明满足条件的在右边
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return lowestCommonAncestor(root.Right,p,q)
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}else {return root}//当前节点的值在给定值的中间(或者等于),即为最深的祖先
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}
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```
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> 普通法
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```go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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//递归会将值层层返回
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func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
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//终止条件
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if root==nil||root.Val==p.Val||root.Val==q.Val{return root}//最后为空或者找到一个值时,就返回这个值
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//后序遍历
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findLeft:=lowestCommonAncestor(root.Left,p,q)
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findRight:=lowestCommonAncestor(root.Right,p,q)
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//处理单层逻辑
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if findLeft!=nil&&findRight!=nil{return root}//说明在root节点的两边
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if findLeft==nil{//左边没找到,就说明在右边找到了
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return findRight
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}else {return findLeft}
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}
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```
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‎problems/0501.二叉搜索树中的众数.md‎

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Original file line numberDiff line numberDiff line change
@@ -428,7 +428,100 @@ class Solution:
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return self.res
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```
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Go:
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暴力法(非BSL)
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```go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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func findMode(root *TreeNode) []int {
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var history map[int]int
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var maxValue int
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var maxIndex int
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var result []int
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history=make(map[int]int)
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traversal(root,history)
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for k,value:=range history{
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if value>maxValue{
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maxValue=value
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maxIndex=k
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}
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}
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for k,value:=range history{
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if value==history[maxIndex]{
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result=append(result,k)
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}
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}
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return result
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}
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func traversal(root *TreeNode,history map[int]int){
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if root.Left!=nil{
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traversal(root.Left,history)
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}
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if value,ok:=history[root.Val];ok{
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history[root.Val]=value+1
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}else{
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history[root.Val]=1
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}
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if root.Right!=nil{
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traversal(root.Right,history)
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}
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}
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```
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计数法BSL(此代码在执行代码里能执行,但提交后报错,不知为何,思路是对的)
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```go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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var count,maxCount int //统计计数
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func findMode(root *TreeNode) []int {
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var result []int
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var pre *TreeNode //前指针
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if root.Left==nil&&root.Right==nil{
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result=append(result,root.Val)
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return result
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}
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traversal(root,&result,pre)
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return result
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}
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func traversal(root *TreeNode,result *[]int,pre *TreeNode){//遍历统计
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//如果BSL中序遍历相邻的两个节点值相同,则统计频率;如果不相同,依据BSL中序遍历排好序的性质,重新计数
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if pre==nil{
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count=1
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}else if pre.Val==root.Val{
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count++
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}else {
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count=1
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}
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//如果统计的频率等于最大频率,则加入结果集;如果统计的频率大于最大频率,更新最大频率且重新将结果加入新的结果集中
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if count==maxCount{
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*result=append(*result,root.Val)
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}else if count>maxCount{
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maxCount=count//重新赋值maxCount
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*result=[]int{}//清空result中的内容
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*result=append(*result,root.Val)
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}
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pre=root//保存上一个的节点
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if root.Left!=nil{
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traversal(root.Left,result,pre)
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}
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if root.Right!=nil{
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traversal(root.Right,result,pre)
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}
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}
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```
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