[pull] master from youngyangyang04:master

problems/0034.在排序数组中查找元素的第一个和最后一个位置.md

@@ -480,6 +480,62 @@ var searchRange = function(nums, target) {
     return [-1, -1];
 };
 ```
+
+
+### TypeScript
+
+```typescript
+function searchRange(nums: number[], target: number): number[] {
+    const leftBoard: number = getLeftBorder(nums, target);
+    const rightBoard: number = getRightBorder(nums, target);
+    // target 在nums区间左侧或右侧
+    if (leftBoard === (nums.length - 1) || rightBoard === 0) return [-1, -1];
+    // target 不存在与nums范围内
+    if (rightBoard - leftBoard <= 1) return [-1, -1];
+    // target 存在于nums范围内
+    return [leftBoard + 1, rightBoard - 1];
+};
+// 查找第一个大于target的元素下标
+function getRightBorder(nums: number[], target: number): number {
+    let left: number = 0,
+        right: number = nums.length - 1;
+    // 0表示target在nums区间的左边
+    let rightBoard: number = 0;
+    while (left <= right) {
+        let mid = Math.floor((left + right) / 2);
+        if (nums[mid] <= target) {
+            // 右边界一定在mid右边（不含mid）
+            left = mid + 1;
+            rightBoard = left;
+        } else {
+            // 右边界在mid左边（含mid）
+            right = mid - 1;
+        }
+    }
+    return rightBoard;
+}
+// 查找第一个小于target的元素下标
+function getLeftBorder(nums: number[], target: number): number {
+    let left: number = 0,
+        right: number = nums.length - 1;
+    // length-1表示target在nums区间的右边
+    let leftBoard: number = nums.length - 1;
+    while (left <= right) {
+        let mid = Math.floor((left + right) / 2);
+        if (nums[mid] >= target) {
+            // 左边界一定在mid左边（不含mid）
+            right = mid - 1;
+            leftBoard = right;
+        } else {
+            // 左边界在mid右边（含mid）
+            left = mid + 1;
+        }
+    }
+    return leftBoard;
+}
+```
+
+
 ### Scala
 ```scala
 object Solution {
@@ -527,5 +583,6 @@ object Solution {
 }
 ```
 
+
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0235.二叉搜索树的最近公共祖先.md

@@ -381,7 +381,36 @@ function lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: Tree
 };
 ```
 
+## Scala
+
+递归:
+
+```scala
+object Solution {
+  def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
+    // scala中每个关键字都有其返回值，于是可以不写return
+    if (root.value > p.value && root.value > q.value) lowestCommonAncestor(root.left, p, q)
+    else if (root.value < p.value && root.value < q.value) lowestCommonAncestor(root.right, p, q)
+    else root
+  }
+}
+```
+
+迭代:
 
+```scala
+object Solution {
+  def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
+    var curNode = root  // 因为root是不可变量，所以要赋值给curNode一个可变量
+    while(curNode != null){
+      if(curNode.value > p.value && curNode.value > q.value) curNode = curNode.left
+      else if(curNode.value < p.value && curNode.value < q.value) curNode = curNode.right
+      else return curNode
+    }
+    null
+  }
+}
+```
 
 
 -----------------------

problems/0236.二叉树的最近公共祖先.md

@@ -343,7 +343,25 @@ function lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: Tree
 };
 ```
 
+## Scala
+
+```scala
+object Solution {
+  def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
+    // 递归结束条件
+    if (root == null || root == p || root == q) {
+      return root
+    }
+
+    var left = lowestCommonAncestor(root.left, p, q)
+    var right = lowestCommonAncestor(root.right, p, q)
 
+    if (left != null && right != null) return root
+    if (left == null) return right
+    left
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0450.删除二叉搜索树中的节点.md

@@ -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>

problems/0701.二叉搜索树中的插入操作.md

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