From 8c694f29966c74e548b70ad48c502ab2f4631923 Mon Sep 17 00:00:00 2001
From: Sainnhe Park <i@sainnhe.dev>
Date: Wed, 1 Feb 2023 10:57:38 +0800
Subject: [PATCH] s0701

---
 ...s0701_insert_into_a_binary_search_tree.hpp | 11 ++++++++
 notes/src/bstree.md                           |  1 +
 ...s0701_insert_into_a_binary_search_tree.cpp | 26 +++++++++++++++++++
 3 files changed, 38 insertions(+)
 create mode 100644 include/s0701_insert_into_a_binary_search_tree.hpp
 create mode 100644 src/s0701_insert_into_a_binary_search_tree.cpp

diff --git a/include/s0701_insert_into_a_binary_search_tree.hpp b/include/s0701_insert_into_a_binary_search_tree.hpp
new file mode 100644
index 0000000..a503062
--- /dev/null
+++ b/include/s0701_insert_into_a_binary_search_tree.hpp
@@ -0,0 +1,11 @@
+#ifndef S0701_INSERT_INTO_A_BINARY_SEARCH_TREE_HPP
+#define S0701_INSERT_INTO_A_BINARY_SEARCH_TREE_HPP
+
+#include "structures.hpp"
+
+class S0701 {
+ public:
+  TreeNode* insertIntoBST(TreeNode* root, int val);
+};
+
+#endif
diff --git a/notes/src/bstree.md b/notes/src/bstree.md
index 1bb9703..d4e75e3 100644
--- a/notes/src/bstree.md
+++ b/notes/src/bstree.md
@@ -1,3 +1,4 @@
 # 二叉搜索树
 
 - [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。
diff --git a/src/s0701_insert_into_a_binary_search_tree.cpp b/src/s0701_insert_into_a_binary_search_tree.cpp
new file mode 100644
index 0000000..e90adea
--- /dev/null
+++ b/src/s0701_insert_into_a_binary_search_tree.cpp
@@ -0,0 +1,26 @@
+#include "s0701_insert_into_a_binary_search_tree.hpp"
+
+TreeNode* S0701::insertIntoBST(TreeNode* root, int val) {
+  TreeNode *cur = root;
+  if (root == nullptr) {
+    return new TreeNode(val);
+  }
+  while (true) {
+    if (val < cur->val) {
+      if (cur->left == nullptr) {
+        cur->left = new TreeNode(val);
+        break;
+      } else {
+        cur = cur->left;
+      }
+    } else {
+      if (cur->right == nullptr) {
+        cur->right = new TreeNode(val);
+        break;
+      } else {
+        cur = cur->right;
+      }
+    }
+  }
+  return root;
+}