Merge pull request #1415 from wzqwtt/tree14

添加(0530.二叉搜索树的最小绝对差、0501.二叉搜索树中的众数) Scala版本
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程序员Carl
2022-07-02 08:48:48 +08:00
committed by GitHub
2 changed files with 149 additions and 1 deletions

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@ -9,7 +9,9 @@
# 501.二叉搜索树中的众数
[力扣题目链接](https://leetcode.cn/problems/find-mode-in-binary-search-tree/solution/)
[力扣题目链接](https://leetcode.cn/problems/find-mode-in-binary-search-tree/)
给定一个有相同值的二叉搜索树BST找出 BST 中的所有众数(出现频率最高的元素)。
@ -798,7 +800,76 @@ function findMode(root: TreeNode | null): number[] {
};
```
## Scala
暴力:
```scala
object Solution {
// 导包
import scala.collection.mutable // 集合包
import scala.util.control.Breaks.{break, breakable} // 流程控制包
def findMode(root: TreeNode): Array[Int] = {
var map = mutable.HashMap[Int, Int]() // 存储节点的值,和该值出现的次数
def searchBST(curNode: TreeNode): Unit = {
if (curNode == null) return
var value = map.getOrElse(curNode.value, 0)
map.put(curNode.value, value + 1)
searchBST(curNode.left)
searchBST(curNode.right)
}
searchBST(root) // 前序遍历把每个节点的值加入到里面
// 将map转换为list随后根据元组的第二个值进行排序
val list = map.toList.sortWith((map1, map2) => {
if (map1._2 > map2._2) true else false
})
var res = mutable.ArrayBuffer[Int]()
res.append(list(0)._1) // 将第一个加入结果集
breakable {
for (i <- 1 until list.size) {
// 如果值相同就加入结果集合反之break
if (list(i)._2 == list(0)._2) res.append(list(i)._1)
else break
}
}
res.toArray // 最终返回res的Array格式return关键字可以省略
}
}
```
递归(利用二叉搜索树的性质):
```scala
object Solution {
import scala.collection.mutable
def findMode(root: TreeNode): Array[Int] = {
var maxCount = 0 // 最大频率
var count = 0 // 统计频率
var pre: TreeNode = null
var result = mutable.ArrayBuffer[Int]()
def searchBST(cur: TreeNode): Unit = {
if (cur == null) return
searchBST(cur.left)
if (pre == null) count = 1 // 等于空置为1
else if (pre.value == cur.value) count += 1 // 与上一个节点的值相同加1
else count = 1 // 与上一个节点的值不同
pre = cur
// 如果和最大值相同,则放入结果集
if (count == maxCount) result.append(cur.value)
// 如果当前计数大于最大值频率,更新最大值,清空结果集
if (count > maxCount) {
maxCount = count
result.clear()
result.append(cur.value)
}
searchBST(cur.right)
}
searchBST(root)
result.toArray // return关键字可以省略
}
}
```
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@ -431,7 +431,84 @@ function getMinimumDifference(root: TreeNode | null): number {
};
```
## Scala
构建二叉树的有序数组:
```scala
object Solution {
import scala.collection.mutable
def getMinimumDifference(root: TreeNode): Int = {
val arr = mutable.ArrayBuffer[Int]()
def traversal(node: TreeNode): Unit = {
if (node == null) return
traversal(node.left)
arr.append(node.value)
traversal(node.right)
}
traversal(root)
// 在有序数组上求最小差值
var result = Int.MaxValue
for (i <- 1 until arr.size) {
result = math.min(result, arr(i) - arr(i - 1))
}
result // 返回最小差值
}
}
```
递归记录前一个节点:
```scala
object Solution {
def getMinimumDifference(root: TreeNode): Int = {
var result = Int.MaxValue // 初始化为最大值
var pre: TreeNode = null // 记录前一个节点
def traversal(cur: TreeNode): Unit = {
if (cur == null) return
traversal(cur.left)
if (pre != null) {
// 对比result与节点之间的差值
result = math.min(result, cur.value - pre.value)
}
pre = cur
traversal(cur.right)
}
traversal(root)
result // return关键字可以省略
}
}
```
迭代解决:
```scala
object Solution {
import scala.collection.mutable
def getMinimumDifference(root: TreeNode): Int = {
var result = Int.MaxValue // 初始化为最大值
var pre: TreeNode = null // 记录前一个节点
var cur = root
var stack = mutable.Stack[TreeNode]()
while (cur != null || !stack.isEmpty) {
if (cur != null) {
stack.push(cur)
cur = cur.left
} else {
cur = stack.pop()
if (pre != null) {
result = math.min(result, cur.value - pre.value)
}
pre = cur
cur = cur.right
}
}
result // return关键字可以省略
}
}
```
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