yan chuanli 2 tahun lalu
induk
melakukan
7230bc52f0
2 mengubah file dengan 137 tambahan dan 10 penghapusan
  1. 27 1
      tree/whole_arrang.py
  2. 110 9
      utils.py

+ 27 - 1
tree/whole_arrang.py

@@ -1,4 +1,30 @@
 from typing import List
+from utils import Bnode, Btree
 
+def permute(nums: List[int]) -> List[List[int]]:
+    node_list = []
+    for i in range(len(List)):
+        node = Bnode(List[i], None, None)
+        node_list.append(node)
 
-def permute(nums: List[int]) -> List[List[int]]:
+    if len(node_list) > 0:
+        for i in range(int(len(List)/2) - 1):
+
+            def array_to_bitree(array):
+                # 判断arr是否为空
+                if len(array) == 0:
+                    return BiTNode(array[0])
+                mid = len(array) // 2  # 有序数组的中间元素的下标
+                # print(mid)
+                # start=0 # 数组第一个元素的下标
+                # end=-1 # 数组最后一个元素的下标
+                if len(array) > 0:
+                    # 将中间元素作为二叉树的根
+                    root = BiTNode(array[mid])
+                    # 如果左边的元素个数不为零,则递归调用函数,生成左子树
+                    if len(array[:mid]) > 0:
+                        root.left_child = arrayToBiTree(array[:mid])
+                    # 如果右边的元素个数不为零,则递归调用函数,生成左子树
+                    if len(array[mid + 1:]) > 0:
+                        root.right_child = arrayToBiTree(array[mid + 1:])
+                return root

+ 110 - 9
utils.py

@@ -1,6 +1,12 @@
 import pytest
 
+# 链表
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
 
+# 数组转链表
 def arr_to_linklist(arr):
 
     p = None
@@ -10,6 +16,7 @@ def arr_to_linklist(arr):
         p = node
     return p
 
+# 链表转数组
 def linkList_to_arr(head):
     res = []
     cur = head
@@ -19,6 +26,7 @@ def linkList_to_arr(head):
 
     return res
 
+# 数组转链表
 def list_to_linklist(arr):
     head = ListNode(arr[0])
     p = head
@@ -27,17 +35,110 @@ def list_to_linklist(arr):
         p = p.next
     return head
 
+# 树
+# class Bnode(object):
+#     def __init__(self, data, left=None, right=None):
+#         self.data = data
+#         self.left = left
+#         self.right = right
+#
+# class Btree(object):
+#     def __init__(self, root=None):
+#         self.root = root
+#
+#     @classmethod
+#     def bulid(cls, list):
+#         node_dict = {}
+#         # 制作节点
+#         for item in list:
+#             data = item['data']
+#             node_dict[data] = Bnode(data)
+#         # 把节点组装成数
+#         for item in list:
+#             data = item['data']
+#             node = node_dict[data]
+#             if node['is_root']:
+#                 root = node
+#             node.left = node_dict.get(item['left'])
+#             node.right = node_dict.get(item['right'])
+#         return cls(root)
+#
+#     def preorder(self, subtree):
+#         # 先序
+#         if subtree is not None:
+#             print(subtree.data)
+#             self.preorder(subtree.left)
+#             self.preorder(subtree.right)
+#
+#     def reverse(self, subtree):
+#         # 后序
+#         if subtree is not None:
+#             subtree.left, subtree.right = subtree.right, subtree.left
+#             self.reverse(self, subtree.left)
+#             self.reverse(self, subtree.right)
 
-class ListNode:
-    def __init__(self, val=0, next=None):
-        self.val = val
-        self.next = next
+#定义二叉树结点类型
+class BiTNode:
+  def __init__(self,root):
+    self.root = root
+    self.left_child = None
+    self.right_child = None
 
+#前序遍历
+def print_tree_pre_order(root):
+    # 先判断二叉树是否为空
+    if root is None:
+        return root
+    # 先根
+    print(root.data)
+    # 再左
+    if root.left_child is not None:
+        print_tree_pre_order(root.left_child)
+    # 再右
+    if root.right_child is not None:
+        print_tree_pre_order(root.right_child)
 
+#中序遍历二叉树
+def print_tree_mid_order(root):
+    # 先判断二叉树是否为空,当左右节点都为空时
+    if root is None:
+        return
+    # 中序遍历 左根右
+    # 遍历左子树
+    if root.left_child is not None:
+        print_tree_mid_order(root.left_child)
+    # 遍历根节点
+    print(root.data)
+    # 遍历右子树
+    if root.right_child is not None:
+        print_tree_mid_order(root.right_child)
 
+#后序遍历
+def print_tree_after_order(root):
+    # 先判断二叉树是否为空
+    if root is None:
+        return root
+    # 再左
+    if root.left_child is not None:
+        print_tree_after_order(root.left_child)
+    # 再右
+    if root.right_child is not None:
+        print_tree_after_order(root.right_child)
+    # 先根
+    print(root.data)
 
-def test_arr_to_linklist():
-    arr = [1, 3, 4, 5, 6]
-    head = arr_to_linklist(arr)
-    arr1 = linkList_to_arr(head)
-    assert  arr == arr1
+def array_to_bitree(array):
+    # 判断arr是否为空
+    if len(array) == 0:
+        return BiTNode(array[0])
+    mid = len(array) // 2  # 有序数组的中间元素的下标
+    if len(array) > 0:
+        # 将中间元素作为二叉树的根
+        root = BiTNode(array[mid])
+        # 如果左边的元素个数不为零,则递归调用函数,生成左子树
+        if len(array[:mid]) > 0:
+            root.left_child = array_to_bitree(array[:mid])
+        # 如果右边的元素个数不为零,则递归调用函数,生成左子树
+        if len(array[mid + 1:]) > 0:
+            root.right_child = array_to_bitree(array[mid + 1:])
+    return root