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添加(0701.二叉搜索树中的插入操作、0450.删除二叉搜索树中的节点) Scala版本 #1418

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Jul 4, 2022
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28 changes: 28 additions & 0 deletions problems/0450.删除二叉搜索树中的节点.md
Original file line number Diff line number Diff line change
Expand Up @@ -582,7 +582,35 @@ function deleteNode(root: TreeNode | null, key: number): TreeNode | null {
};
```

## Scala

```scala
object Solution {
def deleteNode(root: TreeNode, key: Int): TreeNode = {
if (root == null) return root // 第一种情况,没找到删除的节点,遍历到空节点直接返回
if (root.value == key) {
// 第二种情况: 左右孩子都为空,直接删除节点,返回null
if (root.left == null && root.right == null) return null
// 第三种情况: 左孩子为空,右孩子不为空,右孩子补位
else if (root.left == null && root.right != null) return root.right
// 第四种情况: 左孩子不为空,右孩子为空,左孩子补位
else if (root.left != null && root.right == null) return root.left
// 第五种情况: 左右孩子都不为空,将删除节点的左子树头节点(左孩子)放到
// 右子树的最左边节点的左孩子上,返回删除节点的右孩子
else {
var tmp = root.right
while (tmp.left != null) tmp = tmp.left
tmp.left = root.left
return root.right
}
}
if (root.value > key) root.left = deleteNode(root.left, key)
if (root.value < key) root.right = deleteNode(root.right, key)

root // 返回根节点,return关键字可以省略
}
}
```

-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
37 changes: 37 additions & 0 deletions problems/0701.二叉搜索树中的插入操作.md
Original file line number Diff line number Diff line change
Expand Up @@ -585,6 +585,43 @@ function insertIntoBST(root: TreeNode | null, val: number): TreeNode | null {
```


## Scala

递归:

```scala
object Solution {
def insertIntoBST(root: TreeNode, `val`: Int): TreeNode = {
if (root == null) return new TreeNode(`val`)
if (`val` < root.value) root.left = insertIntoBST(root.left, `val`)
else root.right = insertIntoBST(root.right, `val`)
root // 返回根节点
}
}
```

迭代:

```scala
object Solution {
def insertIntoBST(root: TreeNode, `val`: Int): TreeNode = {
if (root == null) {
return new TreeNode(`val`)
}
var parent = root // 记录当前节点的父节点
var curNode = root
while (curNode != null) {
parent = curNode
if(`val` < curNode.value) curNode = curNode.left
else curNode = curNode.right
}
if(`val` < parent.value) parent.left = new TreeNode(`val`)
else parent.right = new TreeNode(`val`)
root // 最终返回根节点
}
}
```


-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>