From 731d1ca8aa61b1b8969ead0c5cf8b7d8b21b55e6 Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Tue, 23 May 2023 21:38:27 -0500
Subject: [PATCH] =?UTF-8?q?Update=200450.=E5=88=A0=E9=99=A4=E4=BA=8C?=
 =?UTF-8?q?=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84=E8=8A=82?=
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---
 .../0450.删除二叉搜索树中的节点.md | 132 +++++++++---------
 1 file changed, 67 insertions(+), 65 deletions(-)

diff --git a/problems/0450.删除二叉搜索树中的节点.md b/problems/0450.删除二叉搜索树中的节点.md
index 2602b528..d4dca1d4 100644
--- a/problems/0450.删除二叉搜索树中的节点.md
+++ b/problems/0450.删除二叉搜索树中的节点.md
@@ -324,88 +324,90 @@ class Solution {
 ```
 
 ## Python 
-
+递归法（版本一）
 ```python
 class Solution:
-    def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
-        if not root : return  None # 节点为空，返回
-        if root.val < key :
-            root.right = self.deleteNode(root.right, key)
-        elif root.val > key :
+    def deleteNode(self, root, key):
+        if root is None:
+            return root
+        if root.val == key:
+            if root.left is None and root.right is None:
+                return None
+            elif root.left is None:
+                return root.right
+            elif root.right is None:
+                return root.left
+            else:
+                cur = root.right
+                while cur.left is not None:
+                    cur = cur.left
+                cur.left = root.left
+                return root.right
+        if root.val > key:
             root.left = self.deleteNode(root.left, key)
-        else:
-            # 当前节点的左子树为空，返回当前的右子树
-            if not root.left : return root.right
-            # 当前节点的右子树为空，返回当前的左子树
-            if not root.right: return root.left
-            # 左右子树都不为空，找到右孩子的最左节点 记为p
-            node = root.right
-            while node.left :
-                node = node.left
-            # 将当前节点的左子树挂在p的左孩子上
-            node.left = root.left
-            # 当前节点的右子树替换掉当前节点，完成当前节点的删除
-            root = root.right
+        if root.val < key:
+            root.right = self.deleteNode(root.right, key)
         return root
 ```
 
-**普通二叉树的删除方式**
+递归法（版本二）
 ```python
 class Solution:
-    def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
-        if not root: return root
-        if root.val == key:
-            if not root.right:  # 这里第二次操作目标值：最终删除的作用
+    def deleteNode(self, root, key):
+        if root is None:  # 如果根节点为空，直接返回
+            return root
+        if root.val == key:  # 找到要删除的节点
+            if root.right is None:  # 如果右子树为空，直接返回左子树作为新的根节点
                 return root.left
-            tmp = root.right
-            while tmp.left:
-                tmp = tmp.left
-            root.val, tmp.val = tmp.val, root.val  # 这里第一次操作目标值：交换目标值其右子树最左面节点。
-            
-        root.left = self.deleteNode(root.left, key)
-        root.right = self.deleteNode(root.right, key)
+            cur = root.right
+            while cur.left:  # 找到右子树中的最左节点
+                cur = cur.left
+            root.val, cur.val = cur.val, root.val  # 将要删除的节点值与最左节点值交换
+        root.left = self.deleteNode(root.left, key)  # 在左子树中递归删除目标节点
+        root.right = self.deleteNode(root.right, key)  # 在右子树中递归删除目标节点
         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
+    def deleteOneNode(self, target: TreeNode) -> TreeNode:
+        """
+        将目标节点（删除节点）的左子树放到目标节点的右子树的最左面节点的左孩子位置上
+        并返回目标节点右孩子为新的根节点
+        是动画里模拟的过程
+        """
+        if target is None:
+            return target
+        if target.right is None:
+            return target.left
+        cur = target.right
+        while cur.left:
+            cur = cur.left
+        cur.left = target.left
+        return target.right
+
+    def deleteNode(self, root: TreeNode, key: int) -> TreeNode:
+        if root is None:
+            return root
+        cur = root
+        pre = None  # 记录cur的父节点，用来删除cur
+        while cur:
+            if cur.val == key:
                 break
+            pre = cur
+            if cur.val > key:
+                cur = cur.left
             else:
-                # Update last and continue
-                last = p
-                if p.val>key:
-                    p = p.left
-                else:
-                    p = p.right
+                cur = cur.right
+        if pre is None:  # 如果搜索树只有头结点
+            return self.deleteOneNode(cur)
+        # pre 要知道是删左孩子还是右孩子
+        if pre.left and pre.left.val == key:
+            pre.left = self.deleteOneNode(cur)
+        if pre.right and pre.right.val == key:
+            pre.right = self.deleteOneNode(cur)
         return root
 ```