From 99eee02c9cc07e681e7f8cf0ee51d9dc7123a633 Mon Sep 17 00:00:00 2001
From: Exzrgs <s.nakao0120@gmail.com>
Date: Wed, 11 Sep 2024 01:00:25 +0900
Subject: [PATCH] complete

---
 .../kth-smallest-element-in-a-bst/step1.py    | 87 +++++++++++++++++++
 .../kth-smallest-element-in-a-bst/step3.py    | 45 ++++++++++
 2 files changed, 132 insertions(+)
 create mode 100644 NeetCode/kth-smallest-element-in-a-bst/step1.py
 create mode 100644 NeetCode/kth-smallest-element-in-a-bst/step3.py

diff --git a/NeetCode/kth-smallest-element-in-a-bst/step1.py b/NeetCode/kth-smallest-element-in-a-bst/step1.py
new file mode 100644
index 0000000..50610a9
--- /dev/null
+++ b/NeetCode/kth-smallest-element-in-a-bst/step1.py
@@ -0,0 +1,87 @@
+"""
+Question:
+Discussion:
+
+なんか再帰関数で管理すべき状態が2つ以上あるときは、返り値にするんじゃなくグローバルな変数として持っておくべき
+そうじゃないと、返り値で何かを管理するとき煩雑になる
+2つ返り値を返すとかも微妙で、返り値に対して条件分岐しないといけなくなる
+更新する答えだけをグローバルとかでもダメ。両方グローバルにする
+    更新した後returnしても再帰関数は終了しないので
+    したのように、count += search(node.left, count) + 1の部分にNoneが返ってしまい、エラーになる
+    
+    class Solution:
+        def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
+            kth_val = 0
+            def search(node, count):
+                if not node:
+                    return count
+                
+                count += search(node.left, count) + 1
+                if count == k:
+                    nonlocal kth_val
+                    kth_val = node.val
+                    return
+                return search(node.right, count)
+            
+            search(root, 0)
+            return kth_val
+
+なので、複数の状態を管理する場合はstackで実装しちゃったほうが良いかもなあ
+値が決まってからreturnしても処理が続いてるのが怖い。勝手に答えを書き換えられないか注意しないといけない
+"""
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
+        self.kth_val = None
+        self.count = 0
+        def search(node):
+            if not node:
+                return
+            
+            search(node.left)
+            self.count += 1
+            if self.count == k:
+                self.kth_val = node.val
+                return
+            search(node.right)
+        
+        search(root)
+        return self.kth_val
+
+# これ良いなあ
+class Solution:
+    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
+        node_stack = []
+        cur = root
+        count = 0
+        while node_stack or cur:
+            while cur:
+                node_stack.append(cur)
+                cur = cur.left
+            cur = node_stack.pop()
+            count += 1
+            if count == k:
+                return cur.val
+            cur = cur.right
+
+# 配列に入れてしまえばcountを管理しなくてよい
+class Solution:
+    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
+        self.sorted_nodes = []
+        
+        def create_sorted_list(node):
+            if not node:
+                return
+            
+            create_sorted_list(node.left)
+            self.sorted_nodes.append(node.val)
+            create_sorted_list(node.right)
+        
+        create_sorted_list(root)
+        return self.sorted_nodes[k - 1]
diff --git a/NeetCode/kth-smallest-element-in-a-bst/step3.py b/NeetCode/kth-smallest-element-in-a-bst/step3.py
new file mode 100644
index 0000000..7a57a31
--- /dev/null
+++ b/NeetCode/kth-smallest-element-in-a-bst/step3.py
@@ -0,0 +1,45 @@
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
+        nodes_stack = []
+        count = 0
+        cur = root
+        while nodes_stack or cur:
+            while cur:
+                nodes_stack.append(cur)
+                cur = cur.left
+            cur = nodes_stack.pop()
+            count += 1
+            if count == k:
+                return cur.val
+            cur = cur.right
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
+        self.smallest = None
+        self.count = 0
+
+        def search(node):
+            if not node:
+                return
+            
+            search(node.left)
+            self.count += 1
+            if self.count == k:
+                self.smallest = node.val
+                return
+            search(node.right)
+        
+        search(root)
+        return self.smallest