feat: generate-parentheses

Author: junbinding

Diff for: recs.generate-parentheses.py

@@ -0,0 +1,31 @@
+from typing import List
+
+
+class Solution:
+    """
+    22. 括号生成
+    https://leetcode-cn.com/problems/generate-parentheses/
+    数字 n 代表生成括号的对数，请你设计一个函数，用于能够生成所有可能的并且 有效的 括号组合。
+    """
+    arr = []
+    def generateParenthesis(self, n: int) -> List[str]:
+        self.arr = []
+        self._gen(0, 0, n * 2, '')
+        return self.arr
+
+
+    def _gen(self, left, right, max, s):
+        # 终止条件
+        if len(s) == max:
+            self.arr.append(s)
+            return s
+
+        # 递归
+        if left < max / 2:
+            self._gen(left + 1, right, max, s + '(')
+        if left > right:
+            self._gen(left, right + 1, max, s + ')')
+
+
+so = Solution()
+print(so.generateParenthesis(1))

Diff for: tree.n-ary-tree-level-order-traversal.py

@@ -0,0 +1,35 @@
+from typing import List
+
+class Node:
+    def __init__(self, val=None, children=None):
+        self.val = val
+        self.children = children
+
+class Solution:
+    """
+    429. N叉树的层序遍历
+    https://leetcode-cn.com/problems/n-ary-tree-level-order-traversal/
+    给定一个 N 叉树，返回其节点值的层序遍历。 (即从左到右，逐层遍历)。
+    """
+    # 迭代
+    def levelOrder(self, root: Node) -> List[List[int]]:
+        if not root:
+            return []
+
+        next_level, res = [root], []
+        i = 0
+        while len(next_level) > 0:
+            tmp_next = []
+            if next_level:
+                res.append([])
+
+            for c in next_level:
+                res[i].append(c.val)
+                if not c.children: continue
+                tmp_next.extend(c.children)
+
+            next_level = tmp_next
+            i += 1
+
+        return res
+