From a0517aa4d00437313061f0c05f3dd0ebc6fdd001 Mon Sep 17 00:00:00 2001
From: Sainnhe Park <i@sainnhe.dev>
Date: Wed, 1 Feb 2023 11:42:09 +0800
Subject: [PATCH] s0450

---
 include/s0450_delete_node_in_a_bst.hpp | 11 ++++++++++
 notes/src/bstree.md                    |  1 +
 src/s0450_delete_node_in_a_bst.cpp     | 30 ++++++++++++++++++++++++++
 3 files changed, 42 insertions(+)
 create mode 100644 include/s0450_delete_node_in_a_bst.hpp
 create mode 100644 src/s0450_delete_node_in_a_bst.cpp

diff --git a/include/s0450_delete_node_in_a_bst.hpp b/include/s0450_delete_node_in_a_bst.hpp
new file mode 100644
index 0000000..6dbd90f
--- /dev/null
+++ b/include/s0450_delete_node_in_a_bst.hpp
@@ -0,0 +1,11 @@
+#ifndef S0450_DELETE_NODE_IN_A_BST_HPP
+#define S0450_DELETE_NODE_IN_A_BST_HPP
+
+#include "structures.hpp"
+
+class S0450 {
+ public:
+  TreeNode* deleteNode(TreeNode* root, int key);
+};
+
+#endif
diff --git a/notes/src/bstree.md b/notes/src/bstree.md
index d4e75e3..dc24e74 100644
--- a/notes/src/bstree.md
+++ b/notes/src/bstree.md
@@ -2,3 +2,4 @@
 
 - [s0235](https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-search-tree/description/): 找两个指定节点的最近公共祖先。思路很简单，只要出现分岔（即一个在当前节点的左边，一个在当前节点的右边），那么这个分岔点就是最近公共祖先。
 - [s0701](https://leetcode.cn/problems/insert-into-a-binary-search-tree/description/): 插入节点。一层一层往下找，直到发现找不到了就在这个地方插入。
+- [s0450](https://leetcode.cn/problems/delete-node-in-a-bst/description/): 删除节点。递归删除。
diff --git a/src/s0450_delete_node_in_a_bst.cpp b/src/s0450_delete_node_in_a_bst.cpp
new file mode 100644
index 0000000..d8e67b3
--- /dev/null
+++ b/src/s0450_delete_node_in_a_bst.cpp
@@ -0,0 +1,30 @@
+#include "s0450_delete_node_in_a_bst.hpp"
+
+TreeNode* S0450::deleteNode(TreeNode* root, int key) {
+  // 如果根节点为空，则直接返回
+  if (root == nullptr) return nullptr;
+  // 如果 key 等于根节点的值，则删除根节点，并将左子树的最右侧节点提上来
+  if (root->val == key) {
+    // 左子树为空的情况
+    if (root->left == nullptr) {
+      return root->right;
+    } else {  // 左子树不为空时
+      // 找到左子树的最右侧节点
+      TreeNode *cur = root->left;
+      while (cur->right) {
+        cur = cur->right;
+      }
+      // 把这个节点提上来，并删除左子树中的该节点
+      cur->left = deleteNode(root->left, cur->val);
+      cur->right = root->right;
+      // 返回
+      return cur;
+    }
+  } else if (key < root->val) {
+    root->left = deleteNode(root->left, key);
+    return root;
+  } else {
+    root->right = deleteNode(root->right, key);
+    return root;
+  }
+}