LeetCode 刷题—树

一：树

1、树描述的是一个父子关系；有节点；根节点；叶子节点三个相关的概念

2、树的高度；深度；层

3、二叉树：每个节点最多只有两个孩子

4、完全二叉树：除了叶子节点；每个孩子并不要求都为两个孩子（从上到下，从左到右依次填满节点）

5、满二叉树：除了叶子节点；每个节点都有两个孩子

6、二叉树的遍历

（1）前序遍历：根节点-->左孩子-->右孩子

（2）中序遍历：左孩子-->根节点-->右孩子

（3）后序遍历：左孩子-->右孩子-->根节点

7、树在程序设计当中；一般会直接给出树的结构；不会让你去创建一个树

二：刷题

144 后序遍历二叉树

(1)思路就是先将数组转化二二叉树；然后根据后序遍历的特性依次访问左子树；右子树；最后访问根节点

from collections import deque
# 定义二叉树节点的类
class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = right
# 迭代方法将列表转换为二叉树
def build_tree_iterative(nodes):if not nodes:return Noneroot = TreeNode(nodes[0])  # 创建根节点queue = deque([root])  # 使用队列来追踪节点index = 1while queue and index < len(nodes):node = queue.popleft()  # 取出队列中的第一个节点# 处理左子树if index < len(nodes) and nodes[index] is not None:node.left = TreeNode(nodes[index])queue.append(node.left)  # 左子节点加入队列index += 1# 处理右子树if index < len(nodes) and nodes[index] is not None:node.right = TreeNode(nodes[index])queue.append(node.right)  # 右子节点加入队列index += 1return root
# 层序遍历打印二叉树的节点值
def print_tree_level_order(root):if not root:print("[]")returnresult = []queue = deque([root])while queue:node = queue.popleft()if node:result.append(node.val)queue.append(node.left)queue.append(node.right)else:result.append(None)# 去除结果列表末尾的 Nonewhile result and result[-1] is None:result.pop()print(result)
# 示例
nodes = [1, None, 2, 3]
root = build_tree_iterative(nodes)
print_tree_level_order(root)

前序遍历树

from collections import deque# 定义二叉树节点的类
class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = right# 迭代方法将列表转换为二叉树
def build_tree_iterative(nodes):if not nodes:return Noneroot = TreeNode(nodes[0])  # 创建根节点queue = deque([root])  # 使用队列来追踪节点index = 1while queue and index < len(nodes):node = queue.popleft()  # 取出队列中的第一个节点# 处理左子树if index < len(nodes) and nodes[index] is not None:node.left = TreeNode(nodes[index])queue.append(node.left)  # 左子节点加入队列index += 1# 处理右子树if index < len(nodes) and nodes[index] is not None:node.right = TreeNode(nodes[index])queue.append(node.right)  # 右子节点加入队列index += 1return root# 前序遍历打印二叉树的节点值
def print_tree_preorder(root):if not root:print("[]")returnresult = []stack = [root]while stack:node = stack.pop()if node:result.append(node.val)stack.append(node.right)  # 先右后左stack.append(node.left)  # 左子节点先入栈，确保左子树优先处理print(result)# 示例
nodes = [1, None, 2, 3]
root = build_tree_iterative(nodes)
print_tree_preorder(root)

中序遍历二叉树

from collections import deque# 定义二叉树节点的类
class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = right# 迭代方法将列表转换为二叉树
def build_tree_iterative(nodes):if not nodes:return Noneroot = TreeNode(nodes[0])  # 创建根节点queue = deque([root])  # 使用队列来追踪节点index = 1while queue and index < len(nodes):node = queue.popleft()  # 取出队列中的第一个节点# 处理左子树if index < len(nodes) and nodes[index] is not None:node.left = TreeNode(nodes[index])queue.append(node.left)  # 左子节点加入队列index += 1# 处理右子树if index < len(nodes) and nodes[index] is not None:node.right = TreeNode(nodes[index])queue.append(node.right)  # 右子节点加入队列index += 1return root# 中序遍历打印二叉树的节点值
def print_tree_inorder(root):if not root:print("[]")returnresult = []stack = []current = rootwhile stack or current:# 遍历左子树while current:stack.append(current)current = current.left# 处理当前节点current = stack.pop()result.append(current.val)# 遍历右子树current = current.rightprint(result)# 示例
nodes = [1, None, 2, 3]
root = build_tree_iterative(nodes)
print_tree_inorder(root)