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添加0071.0501.二叉搜索树中的众数 go版本
添加0071.0501.二叉搜索树中的众数 go版本
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@ -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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