Merge pull request youngyangyang04#1693 from roylx/master

0701.二叉搜索树中的插入操作 - 优化了Python3 迭代法 和 增加了 Python3 递归法 - 无返回值 0450.删除二叉搜索树中的节点 - 添加python3迭代法

Author: youngyangyang04

problems/0450.删除二叉搜索树中的节点.md

@@ -344,6 +344,48 @@ class Solution:
         return root
 ```
 
+**迭代法**
+```python
+class Solution:
+    def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
+        # 找到节点后分两步，1. 把节点的左子树和右子树连起来，2. 把右子树跟父节点连起来
+        # root is None
+        if not root: return root
+        p = root
+        last = None
+        while p:
+            if p.val==key:
+                # 1. connect left to right
+                # right is not None -> left is None | left is not None
+                if p.right:
+                    if p.left:
+                        node = p.right
+                        while node.left:
+                            node = node.left
+                        node.left = p.left
+                    right = p.right
+                else:
+                # right is None -> right=left
+                    right = p.left
+                # 2. connect right to last
+                if last==None:
+                    root = right
+                elif last.val>key:
+                    last.left = right
+                else:
+                    last.right = right
+                # 3. return
+                break
+            else:
+                # Update last and continue
+                last = p
+                if p.val>key:
+                    p = p.left
+                else:
+                    p = p.right
+        return root
+```
+
 ## Go
 ```Go
 // 递归版本

problems/0701.二叉搜索树中的插入操作.md

@@ -330,23 +330,42 @@ class Solution:
         return root
 ```
 
+**递归法** - 无返回值 有注释 不用Helper function
+```python
+class Solution:
+    def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
+        if not root:    # for root==None
+            return TreeNode(val)
+        if root.val<val:
+            if root.right==None:    # find the parent
+                root.right = TreeNode(val)
+            else:   # not found, keep searching
+                self.insertIntoBST(root.right, val)
+        if root.val>val:
+            if root.left==None: # found the parent
+                root.left = TreeNode(val)
+            else:   # not found, keep searching
+                self.insertIntoBST(root.left, val)
+        # return the final tree
+        return root
+```
+
 **迭代法**  
 与无返回值的递归函数的思路大体一致 
 ```python
 class Solution:
     def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
         if not root: 
             return TreeNode(val)
-        parent = None
+        parent = None # 此步可以省略
         cur = root
 
         # 用while循环不断地找新节点的parent
         while cur: 
+            parent = cur  # 首先保存当前非空节点作为下一次迭代的父节点
             if cur.val < val: 
-                parent = cur
                 cur = cur.right
             elif cur.val > val: 
-                parent = cur
                 cur = cur.left
 
         # 运行到这意味着已经跳出上面的while循环,