feat(leetcode; utils): 94.二叉树的中序遍历

94.二叉树的中序遍历

Signed-off-by: Ching <loooching@gmail.com>

.gitignore

@@ -86,7 +86,7 @@ ipython_config.py
 # pyenv
 #   For a library or package, you might want to ignore these files since the code is
 #   intended to run in multiple environments; otherwise, check them in:
-# .python-version
+.python-version
 
 # pipenv
 #   According to pypa/pipenv#598, it is recommended to include Pipfile.lock in version control.

.vscode/launch.json

@@ -0,0 +1,15 @@
+{
+  // 使用 IntelliSense 了解相关属性。 
+  // 悬停以查看现有属性的描述。
+  // 欲了解更多信息，请访问: https://go.microsoft.com/fwlink/?linkid=830387
+  "version": "0.2.0",
+  "configurations": [
+    {
+      "name": "Python: 当前文件",
+      "type": "python",
+      "request": "launch",
+      "program": "${file}",
+      "console": "integratedTerminal"
+    }
+  ]
+}

.vscode/settings.json

@@ -0,0 +1,17 @@
+{
+  "python.formatting.provider": "black",
+  "python.formatting.blackArgs": [
+    "-S",
+    "-C",
+    "-l 120"
+  ],
+  "editor.formatOnPaste": true,
+  "editor.formatOnSave": true,
+  "editor.rulers": [
+    120
+  ],
+  "editor.bracketPairColorization.enabled": true,
+  "editor.guides.bracketPairs": "active",
+  "vetur.format.options.tabSize": 4,
+  "vetur.format.enable": false
+}

94.二叉树的中序遍历.py

@@ -0,0 +1,57 @@
+#
+# @lc app=leetcode.cn id=94 lang=python3
+#
+# [94] 二叉树的中序遍历
+#
+# 中序遍历（LDR）是二叉树遍历的一种，也叫做中根遍历、中序周游。在二叉树中，
+# 中序遍历首先遍历左子树，然后访问根结点，最后遍历右子树。
+
+# @lc code=start
+# 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
+
+from typing import Optional, List
+
+from utils import *
+
+
+class Solution:
+    def inorderTraversal(self, root: TreeNode) -> List[int]:
+        res = []
+        stack = []
+        if root:
+            stack.append(root)
+        while stack:
+            node = stack[-1]
+            if node:
+                # 将该节点弹出，避免重复操作，下面再将右中左节点添加到栈中
+                stack.pop()
+                # 添加右节点（空节点不入栈）
+                if node.right:
+                    stack.append(node.right)
+                # 添加中节点
+                stack.append(node)
+                # 中节点访问过，但是还没有处理，加入空节点作为标记
+                stack.append(None)
+                # 添加左节点（空节点不入栈）
+                if node.left:
+                    stack.append(node.left)
+            else:  # 只有在遇到空节点的时候，才将下一个节点放进结果集中
+                # 将空节点弹出
+                stack.pop()
+                # 重新取出栈中元素
+                node = stack.pop()
+                res.append(node.val)
+        return res
+
+
+s = Solution()
+root = creatBTree([1, None, 2], 0)
+s.inorderTraversal(root)
+
+
+# @lc code=end

utils/__init__.py

@@ -0,0 +1,63 @@
+# -*- coding: UTF-8 -*-
+
+
+class TreeNode(object):
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+    def __repr__(self):
+        return 'node %s' % self.val
+
+
+def creatBTree(data, index):
+    """通过数组生成二叉树
+
+    Args:
+        data ([int, None]): 每个节点都要定义，包括空节点 eg. [1,None,2, None, None, 3]
+        index (int): 返回哪个位置的节点
+
+    Returns:
+        TreeNode: TreeNode obj
+    """
+    pNode = None
+    if index < len(data):
+        if data[index] == None:
+            return
+        pNode = TreeNode(data[index])
+        pNode.left = creatBTree(data, 2 * index + 1)  # [1, 3, 7, 15, ...]
+        pNode.right = creatBTree(data, 2 * index + 2)  # [2, 5, 12, 25, ...]
+    return pNode
+
+
+def generate_tree(vals):
+    """通过 leetcode 样式的数组生成二叉树
+
+    Args:
+        vals ([int, None]): leetcode 格式，省略全空子节点 eg. [1, None, 2, 3]
+
+    Returns:
+        TreeNode: TreeNode obj
+    """
+    if not vals:
+        return None
+    que = []
+    fill_left = True
+    for val in vals:
+        node = TreeNode(val) if val else None
+        if len(que) == 0:
+            root = node
+            que.append(node)
+        elif fill_left:
+            que[0].left = node
+            fill_left = False
+            if node:
+                que.append(node)
+        else:
+            que[0].right = node
+            if node:
+                que.append(node)
+            que.pop(0)
+            fill_left = True
+    return root