本题是二叉树的层序遍历模板：每次循环将一层节点出队，再将一层节点入队，也是所有可用层序遍历解二叉树题目的模板，只需要在模板里稍加改动即可解题

from typing import List, Optional, Union
from collections import deque
'''
102. 二叉树的层序遍历
给你二叉树的根节点 root ，返回其节点值的 层序遍历 。 （即逐层地，从左到右访问所有节点）。
示例 1:输入：root = [3,9,20,null,null,15,7]输出：[[3],[9,20],[15,7]]
题眼：基础
思路：利用队列实现
'''class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef buildBinaryTree(nums: List[Union[int, str]]) -> Optional[TreeNode]:if len(nums) == 0:return Noneif len(nums) == 1:return TreeNode(nums[0])# 1、转换为顺序存储tmpTree = []for n in nums:if n != 'null':tmpTree.append(TreeNode(n))else:tmpTree.append(None)# 2、转换为链式存储：双指针root = tmpTree[0]parent, child = 0, 1while child < len(tmpTree):if tmpTree[parent] != None:  # 双亲节点有效if tmpTree[child] != None:  # 左孩子节点有效tmpTree[parent].left = tmpTree[child]child += 1  # 指向右孩子if child < len(tmpTree) and tmpTree[child] != None:  # 右孩子节点存在且有效tmpTree[parent].right = tmpTree[child]child += 1  # 指向下个节点的左孩子parent += 1  # 更新遍历双亲节点return rootclass Solution:def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:if root == None:return []result = []que = deque()que.append(root)  # 队列初始值while len(que) > 0:size = len(que)  # 当前队列长度==二叉树一层的节点个数layer = []for _ in range(size):  # 一层遍历node = que.popleft()  # 记录下出队的节点layer.append(node.val)# 和之前深度遍历一样：空节点不入队，不然对None节点取值会出错if node.left != None:que.append(node.left)if node.right != None:que.append(node.right)result.append(layer)return resultif __name__ == "__main__":obj = Solution()while True:try:in_line = input().strip().split('=')[1].strip()[1: -1]nums = []if in_line != '':for n in in_line.split(','):if n != 'null':nums.append(int(n))else:nums.append(n)root = buildBinaryTree(nums)print(obj.levelOrder(root))except EOFError:break

“102. 二叉树的层序遍历”的结果反转/逆序即可

from typing import List, Optional, Union
from collections import deque
'''
107. 二叉树的层序遍历 II
给你二叉树的根节点 root ，返回其节点值 自底向上的层序遍历 。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）
示例 1:输入：root = [3,9,20,null,null,15,7]输出：[[15,7],[9,20],[3]]
题眼：自底向上的层序遍历
思路：“102. 二叉树的层序遍历”的结果反转/逆序即可
'''class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef buildBinaryTree(nums: List[Union[int, str]]) -> Optional[TreeNode]:if len(nums) == 0:return Noneif len(nums) == 1:return TreeNode(nums[0])# 1、顺序存储tmpTree = []for n in nums:if n != 'null':tmpTree.append(TreeNode(n))else:tmpTree.append(None)# 2、链式存储：双指针root = tmpTree[0]parent, child = 0, 1while child < len(tmpTree):if tmpTree[parent] != None:  # 双亲节点有效if tmpTree[child] != None:  # 左孩子有效tmpTree[parent].left = tmpTree[child]child += 1  # 指向右孩子if child < len(tmpTree) and tmpTree[child] != None:  # 右孩子存在且有效tmpTree[parent].right = tmpTree[child]child += 1  # 指向下个节点的左孩子parent += 1return rootclass Solution:def levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:# 情况1、树为空if root == None:return []# 情况2、树不为空que = deque()que.append(root)result = []while len(que) > 0:size = len(que)  # 当前队列长度==二叉树一层的节点个数layer = []for _ in range(size):  # 一层遍历node = que.popleft()layer.append(node.val)if node.left != None:que.append(node.left)if node.right != None:que.append(node.right)result.append(layer)# 反转result# left, right = 0, len(result) - 1# while left < right:#     result[left], result[right] = result[right], result[left]#     left += 1#     right -= 1return result[::-1]if __name__ == "__main__":obj = Solution()while True:try:in_line = input().strip().split('=')[1].strip()[1: -1]nums = []if in_line != '':for n in in_line.split(','):if n != 'null':nums.append(int(n))else:nums.append(n)print(nums)root = buildBinaryTree(nums)print(obj.levelOrderBottom(root))except EOFError:break

一般 层序遍历的基础上，要访问每个节点的N个孩子

from typing import List, Optional, Union
from collections import deque
'''
429. N 叉树的层序遍历
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）。
树的序列化输入是用层序遍历，每组子节点都由 null 值分隔（参见示例）。
示例 1:输入：root = [1,null,3,2,4,null,5,6]输出：[[1],[3,2,4],[5,6]]
题眼：N 叉树的层序遍历
思路：一般 层序遍历的基础上，要访问每个节点的N个孩子
'''class Node:def __init__(self, val=None, children=None):self.val = valself.children = childrenclass Solution:def levelOrder(self, root: Node) -> List[List[int]]:# 情况1、树为空if root == None:return []# 情况2、树不为空que = deque()que.append(root)result = []while len(que) > 0:size = len(que)layer = []for _ in range(size):  # 一层遍历node = que.popleft()layer.append(node.val)if node.children != None:for child in node.children:que.append(child)result.append(layer)return result

from typing import List, Optional, Union
from collections import deque
'''
637. 二叉树的层平均值
给定一个非空二叉树的根节点 root , 以数组的形式返回每一层节点的平均值。与实际答案相差 10-5 以内的答案可以被接受。
示例 1:输入：root = [3,9,20,null,null,15,7]输出：[3.00000,14.50000,11.00000]解释：第 0 层的平均值为 3,第 1 层的平均值为 14.5,第 2 层的平均值为 11 。因此返回 [3, 14.5, 11] 。
题眼：每一层节点的平均值
思路：一般 层序遍历的基础上，计算每一层对应的平均值
'''class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef buildBinaryTree(nums: List[Union[int, str]]) -> Optional[TreeNode]:if len(nums) == 0:return Noneif len(nums) == 1:return TreeNode(nums[0])# 1、顺序存储tmpTree = []for n in nums:if n != 'null':tmpTree.append(TreeNode(n))else:tmpTree.append(None)# 2、转换为链式存储root = tmpTree[0]parent, child = 0, 1while child < len(tmpTree):if tmpTree[parent] != None:  # 双亲节点有效if tmpTree[child] != None:  # 左孩子节点有效tmpTree[parent].left = tmpTree[child]child += 1  # 指向右孩子if child < len(tmpTree) and tmpTree[child] != None:  # 右孩子存在且有效tmpTree[parent].right = tmpTree[child]child += 1  # 指向下个节点的左孩子parent += 1return rootclass Solution:def averageOfLevels(self, root: Optional[TreeNode]) -> List[float]:que = deque()que.append(root)result = []while len(que) > 0:size = len(que)  # 当前队列长度==二叉树一层的节点个数s = 0for _ in range(size):  # 一层遍历node = que.popleft()s += node.valif node.left != None:que.append(node.left)if node.right != None:que.append(node.right)result.append(s / size)return resultif __name__ == "__main__":obj = Solution()while True:try:in_line = input().strip().split('=')[1].strip()[1: -1]nums = []if in_line != '':for n in in_line.split(','):if n != 'null':nums.append(int(n))else:nums.append(n)root = buildBinaryTree(nums)print(obj.averageOfLevels(root))except EOFError:break

from typing import List, Optional, Union
from collections import deque
'''
515. 在每个树行中找最大值
给定一棵二叉树的根节点 root ，请找出该二叉树中每一层的最大值。
示例 1:输入: root = [1,3,2,5,3,null,9]输出: [1,3,9]
题眼：二叉树的层序遍历
思路：一般 层序遍历的基础上，记录每一行的最大值
'''class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef buildBinaryTree(nums: List[Union[int, str]]) -> Optional[TreeNode]:if len(nums) == 0:return Noneif len(nums) == 1:return TreeNode(nums[0])# 1、转换为顺序存储tmpTree = []for n in nums:if n != 'null':tmpTree.append(TreeNode(n))else:tmpTree.append(None)root = tmpTree[0]# 2、转换为链式存储：双指针parent, child = 0, 1while child < len(tmpTree):if tmpTree[parent] != None:  # 双亲节点有效if tmpTree[child] != None:  #  左孩子有效tmpTree[parent].left = tmpTree[child]child += 1  # 指向右孩子if child < len(tmpTree) and tmpTree[child] != None:  # 右孩子存在且有效tmpTree[parent].right = tmpTree[child]child += 1  # 指向下个节点的左孩子parent += 1return rootclass Solution:def largestValues(self, root: Optional[TreeNode]) -> List[int]:# 情况1、树为空if root == None:return []# 情况2、树不为空que = deque()que.append(root)result = []while len(que) > 0:size = len(que)  # 当前队列长度==二叉树一层的节点个数n = -float('inf')for _ in range(size):  # 一层遍历node = que.popleft()n = max(n, node.val)if node.left != None:que.append(node.left)if node.right != None:que.append(node.right)result.append(n)return resultif __name__ == "__main__":obj = Solution()while True:try:in_line = input().strip().split('=')[1].strip()[1: -1]nums = []if in_line != '':for n in in_line.split(','):if n != 'null':nums.append(int(n))else:nums.append(n)root = buildBinaryTree(nums)print(obj.largestValues(root))except EOFError:break

一般 层序遍历的基础上，返回每一层的最后一个元素

from typing import List, Optional, Union
from collections import deque
'''
199. 二叉树的右视图
给定一个二叉树的 根节点 root，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。
示例 1:输入：[1,2,3,null,5,null,4]输出：[1,3,4]
题眼：从顶部到底部  从右侧所能看到
思路：一般 层序遍历的基础上，返回每一层的最后一个元素
'''class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef buildBinaryTree(nums: List[Union[int, str]]) -> Optional[TreeNode]:if len(nums) == 0:return Noneif len(nums) == 1:return TreeNode(nums[0])# 1、 顺序存储tmpTree = []for n in nums:if n != 'null':tmpTree.append(TreeNode(n))else:tmpTree.append(None)# 2、链式存储：双指针root = tmpTree[0]parent, child = 0, 1while child < len(tmpTree):if tmpTree[parent] != None:  # 双亲结点有效if tmpTree[child] != None:  # 左孩子节点有效tmpTree[parent].left = tmpTree[child]child += 1  # 指向右孩子if child < len(tmpTree) and tmpTree[child] != None:  # 右孩子节点存在且有效tmpTree[parent].right = tmpTree[child]child += 1  # 指向下个节点的左孩子parent += 1return rootclass Solution:def rightSideView(self, root: Optional[TreeNode]) -> List[int]:# 情况1、树为空if root == None:return []# 情况2、树不为空que = deque()que.append(root)result = []while len(que) > 0:size = len(que)for i in range(size):  # 一层遍历node = que.popleft()if i == size - 1:  # 添加每一层的最后一个元素result.append(node.val)if node.left != None:que.append(node.left)if node.right != None:que.append(node.right)return resultif __name__ == "__main__":obj = Solution()while True:try:in_line = input().strip()[1: -1]nums = []if in_line != '':for n in in_line.split(','):if n != 'null':nums.append(int(n))else:nums.append(n)root = buildBinaryTree(nums)print(obj.rightSideView(root))except EOFError:break

一般 层序遍历的基础上，访问每一层的每个元素时，都把第一个元素进行赋值即可

from typing import List, Optional, Union
from collections import deque
'''
513. 找树左下角的值
给定一个二叉树的 根节点 root，请找出该二叉树的 最底层 最左边 节点的值。
假设二叉树中至少有一个节点。
示例 1:输入: root = [2,1,3]输出: 1
题眼：二叉树的层序遍历
思路：一般 层序遍历的基础上，访问每一层的每个元素时，都把第一个元素进行赋值即可
'''class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef buildBinaryTree(nums: List[Union[int, str]]) -> Optional[TreeNode]:if len(nums) == 0:return Noneif len(nums) == 1:return TreeNode(nums[0])# 1、顺序存储tmpTree = []for n in nums:if n != 'null':tmpTree.append(TreeNode(n))else:tmpTree.append(None)root = tmpTree[0]# 2、链式存储parent, child = 0, 1while child < len(tmpTree):if tmpTree[parent] != None:  # 双亲结点有效if tmpTree[child] != None:  # 左孩子节点有效tmpTree[parent].left = tmpTree[child]child += 1  # 指向右孩子if child < len(tmpTree) and tmpTree[child] != None:  # 右孩子存在且有效tmpTree[parent].right = tmpTree[child]child += 1  # 指向下一个节点的左孩子parent += 1return rootclass Solution:def findBottomLeftValue(self, root: Optional[TreeNode]) -> int:que = deque()que.append(root)result = 0while len(que) > 0:size = len(que)for i in range(size):node = que.popleft()if i == 0:  # 总是获取层序遍历的每一层第一个值result = node.valif node.left != None:que.append(node.left)if node.right != None:que.append(node.right)return resultif __name__ == "__main__":obj = Solution()while True:try:in_line = input().strip().split('=')[1].strip()[1: -1]nums = []if in_line != '':for n in in_line.split(','):if n != 'null':nums.append(int(n))else:nums.append(n)root = buildBinaryTree(nums)print(obj.findBottomLeftValue(root))except EOFError:break

一般 层序遍历的基础上，对结果的偶数个逆序一下

from typing import List, Optional, Union
from collections import deque
'''
103. 二叉树的锯齿形层序遍历
给你二叉树的根节点 root ，返回其节点值的 锯齿形层序遍历 。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。
示例 1:输入：root = [3,9,20,null,null,15,7]输出：[[3],[20,9],[15,7]]
题眼：基础
思路：一般 层序遍历的基础上，对结果的偶数个逆序一下
'''class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef buildBinaryTree(nums: List[Union[int, str]]) -> Optional[TreeNode]:if len(nums) == 0:return Noneif len(nums) == 1:return TreeNode(nums[0])# 1、转换为顺序存储tmpTree = []for n in nums:if n != 'null':tmpTree.append(TreeNode(n))else:tmpTree.append(None)# 2、转换为链式存储：双指针root = tmpTree[0]parent, child = 0, 1while child < len(tmpTree):if tmpTree[parent] != None:  # 双亲节点有效if tmpTree[child] != None:  # 左孩子节点有效tmpTree[parent].left = tmpTree[child]child += 1  # 指向右孩子if child < len(tmpTree) and tmpTree[child] != None:  # 右孩子节点存在且有效tmpTree[parent].right = tmpTree[child]child += 1  # 指向下个节点的左孩子parent += 1  # 更新遍历双亲节点return rootclass Solution:def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:# 情况1、树为空if root == None:return []# 情况2、树不为空que = deque()que.append(root)result = []while len(que) > 0:size = len(que)layer = []for _ in range(size):node = que.popleft()layer.append(node.val)if node.left != None:que.append(node.left)if node.right != None:que.append(node.right)result.append(layer)# 对结果的偶数个逆序一下for i in range(1, len(result), 2):result[i] = result[i][::-1]return resultif __name__ == "__main__":obj = Solution()while True:try:in_line = input().strip().split('=')[1].strip()[1: -1]nums = []if in_line != '':for n in in_line.split(','):if n != 'null':nums.append(int(n))else:nums.append(n)root = buildBinaryTree(nums)print(obj.zigzagLevelOrder(root))except EOFError:break

一般 层序遍历的基础上，将每层的节点（除最后一个外）的next值指向下一个节点

from typing import List, Optional, Union
from collections import deque
'''
116. 填充每个节点的下一个右侧节点指针
给定一个 完美二叉树 ，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：
struct Node {int val;Node *left;Node *right;Node *next;
}
填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。
初始状态下，所有next 指针都被设置为 NULL。
示例 1:输入：root = [1,2,3,4,5,6,7]输出：[1,#,2,3,#,4,5,6,7,#]解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。序列化的输出按层序遍历排列，同一层节点由 next 指针连接，'#' 标志着每一层的结束。
题眼：二叉树的层序遍历（满二叉树）
思路：一般 层序遍历的基础上，将每层的节点（除最后一个外）的next值指向下一个节点
'''class Node:def __init__(self, val=0, left=None, right=None, next=None):self.val = valself.left = leftself.right = rightself.next = nextdef buildBinartTree(nums: List[Union[int]]) -> Optional[Node]:if len(nums) == 0:return []if len(nums) == 1:return Node(nums[0])# 1、转换为顺序存储tmpTree = []for n in nums:tmpTree.append(Node(n))root = tmpTree[0]# 2、转换为链式存储parent, child = 0, 1while child < len(tmpTree):tmpTree[parent].left = tmpTree[child]  # 满二叉树左孩子一定有效child += 1tmpTree[parent].right = tmpTree[child]  # 满二叉树右孩子一定存在且有效child += 1parent += 1return rootclass Solution:def connect(self, root: Optional[Node]) -> Optional[Node]:# 情况1、树为空if root == None:return root# 情况2、树不为空que = deque()que.append(root)while len(que) > 0:size = len(que)pre = Nonefor i in range(size):node = que.popleft()if i == 0:  # 记录每层第一个节点为prepre = nodeelse:  # 从每层第二个节点开始，都要给pre的next指针赋值pre.next = nodepre = nodeif node.left != None:que.append(node.left)if node.right != None:que.append(node.right)return rootif __name__ == "__main__":while True:try:in_line = input().strip().split('=')[1].strip()[1: -1]nums = []if in_line != '':for n in in_line.split(','):nums.append(int(n))print(nums)except EOFError:break

一般 层序遍历的基础上，将每层的节点（除最后一个外）的next值指向下一个节点

from typing import List, Union, Optional
from collections import deque
'''
117. 填充每个节点的下一个右侧节点指针 II
给定一个二叉树
struct Node {int val;Node *left;Node *right;Node *next;
}
填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。
初始状态下，所有next 指针都被设置为 NULL。
示例 1:输入：root = [1,2,3,4,5,null,7]输出：[1,#,2,3,#,4,5,7,#]解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。序列化输出按层序遍历顺序（由 next 指针连接），'#' 表示每层的末尾。
题眼：二叉树的层序遍历（注意不是满二叉树）
思路：一般 层序遍历的基础上，将每层的节点（除最后一个外）的next值指向下一个节点
'''class Node:def __init__(self, val=0, left=None, right=None, next=None):self.val = valself.left = leftself.right = rightself.next = nextdef buildBinaryTree(nums: List[Union[int, str]]) -> Optional[Node]:if len(nums) == 0:return Noneif len(nums) == 1:return Node(nums[0])# 1、转换为顺序存储tmpTree = []for n in nums:if n != 'null':tmpTree.append(Node(n))else:tmpTree.append(None)root = tmpTree[0]# 2、转换为链式存储parent, child = 0, 1while child < len(tmpTree):if tmpTree[parent] != None:  # 双亲节点有效性判断if tmpTree[child] != None:  # 左孩子节点有效性判断tmpTree[parent].left = tmpTree[child]child += 1if child < len(tmpTree) and tmpTree[child] != None:  # 左孩子节点存在且有效tmpTree[parent].right = tmpTree[child]child += 1parent += 1return rootclass Solution:def connect(self, root: Optional[Node]) -> Optional[Node]:# 情况1、树为空if root == None:return root# 情况2、树不为空que = deque()que.append(root)while len(que) > 0:size = len(que)pre = Nonefor i in range(size):node = que.popleft()if i == 0:  # 记录每层第一个节点为prepre = nodeelse:  # 从每层第二个节点开始，都要给pre的next指针赋值pre.next = nodepre = nodeif node.left != None:que.append(node.left)if node.right != None:que.append(node.right)return rootif __name__ == "__main__":while True:try:in_line = input().strip().split('=')[1].strip()[1: -1]nums = []if in_line != '':for n in in_line.split(','):if n != 'null':nums.append(int(n))else:nums.append(n)print(nums)except EOFError:break

一般 层序遍历的基础上，输出最大高度值，也就是二叉树的高度值

from typing import List, Optional, Union
from collections import deque
'''
104. 二叉树的最大深度
给定一个二叉树，找出其最大深度。
二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。
说明: 叶子节点是指没有子节点的节点。
示例 1:输入：[3,9,20,null,null,15,7]输出：3
题眼：二叉树的层序遍历
思路：一般 层序遍历的基础上，输出最大高度值
'''class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef buildBinaryTree(nums: List[Union[int, str]]) -> Optional[TreeNode]:if len(nums) == 0:return Noneif len(nums) == 1:return TreeNode(nums[0])# 1、顺序存储tmpTree = []for n in nums:if n != 'null':tmpTree.append(TreeNode(n))else:tmpTree.append(None)root = tmpTree[0]# 2、链式存储parent, child = 0, 1while child < len(tmpTree):if tmpTree[parent] != None:  # 双亲结点有效if tmpTree[child] != None:  # 左孩子节点有效tmpTree[parent].left = tmpTree[child]child += 1  # 指向右孩子if child < len(tmpTree) and tmpTree[child] != None:  # 右孩子存在且有效tmpTree[parent].right = tmpTree[child]child += 1  # 指向下一个节点的左孩子parent += 1return rootclass Solution:def maxDepth(self, root: Optional[TreeNode]) -> int:# 情况1、树为空if root == None:return 0# 情况2、树不为空result = 0que = deque()que.append(root)while len(que) > 0:size = len(que)for _ in range(size):node = que.popleft()if node.left != None:que.append(node.left)if node.right != None:que.append(node.right)result += 1return result# 做完了计算完全二叉树的节点数，用了带return的递归法，有点感觉了，看看能否把这里的迭代法写出来def maxDepthRecursive(self, root: Optional[TreeNode]) -> int:# 1、确定函数参数和返回值def maxDepth(node: Optional[TreeNode]) -> int:# 2、终止条件if node == None:return 0# 3、单次递归的操作leftN = maxDepth(node.left)     # 左rightN = maxDepth(node.right)   # 右return max(leftN, rightN) + 1   # 中return maxDepth(root)if __name__ == "__main__":obj = Solution()while True:try:in_line = input().strip().split('=')[1].strip()[1: -1]nums = []if in_line != '':for n in in_line.split(','):if n != 'null':nums.append(int(n))else:nums.append(n)root = buildBinaryTree(nums)print(obj.maxDepth(root))except EOFError:break

一般 层序遍历的基础上，输出最小高度值（第一次遍历到叶子节点的时候）

from typing import List, Optional, Union
from collections import deque
'''
111. 二叉树的最小深度
给定一个二叉树，找出其最小深度。
最小深度是从根节点到最近叶子节点的最短路径上的节点数量。
说明：叶子节点是指没有子节点的节点。
示例 1:输入：root = [3,9,20,null,null,15,7]输出：2
题眼：二叉树的层序遍历
思路：一般 层序遍历的基础上，输出最小高度值（第一次遍历到叶子节点的时候）
'''class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef buildBinaryTree(nums: List[Union[int, str]]) -> Optional[TreeNode]:if len(nums) == 0:return Noneif len(nums) == 1:return TreeNode(nums[0])# 1、顺序存储tmpTree = []for n in nums:if n != 'null':tmpTree.append(TreeNode(n))else:tmpTree.append(None)root = tmpTree[0]# 2、链式存储parent, child = 0, 1while child < len(tmpTree):if tmpTree[parent] != None:  # 双亲结点有效if tmpTree[child] != None:  # 左孩子节点有效tmpTree[parent].left = tmpTree[child]child += 1  # 指向右孩子if child < len(tmpTree) and tmpTree[child] != None:  # 右孩子存在且有效tmpTree[parent].right = tmpTree[child]child += 1  # 指向下一个节点的左孩子parent += 1return rootclass Solution:def minDepth(self, root: Optional[TreeNode]) -> int:# 情况1、树为空if root == None:return 0# 情况2、树不为空result = 0que = deque()que.append(root)while len(que) > 0:size = len(que)for _ in range(size):node = que.popleft()if node.left == None and node.right == None:return result + 1if node.left != None:que.append(node.left)if node.right != None:que.append(node.right)result += 1return result# 做完了最大深度的递归，试试这个最小深度的，有陷阱！！！def minDepthRecursive(self, root: Optional[TreeNode]) -> int:# 1、确定函数参数和返回值def minD(node: Optional[TreeNode]) -> int:# 2、终止条件if node == None:return 0# 3、单次递归的操作leftN = minD(node.left)                         # 左rightN = minD(node.right)                       # 右# 当一个左子树为空，右不为空，这时并不是最低点if node.left == None and node.right != None:    # 中return 1 + rightN# 当一个右子树为空，左不为空，这时并不是最低点if node.left != None and node.right == None:return 1 + leftNreturn min(leftN, rightN) + 1return minD(root)if __name__ == "__main__":obj = Solution()while True:try:in_line = input().strip().split('=')[1].strip()[1: -1]nums = []if in_line != '':for n in in_line.split(','):if n != 'null':nums.append(int(n))else:nums.append(n)root = buildBinaryTree(nums)print(obj.minDepth(root))except EOFError:break

一般 层序遍历的基础上，遍历children部分

from typing import List, Optional, Union
from collections import deque
'''
559. N 叉树的最大深度
给定一个 N 叉树，找到其最大深度。
最大深度是指从根节点到最远叶子节点的最长路径上的节点总数。
N 叉树输入按层序遍历序列化表示，每组子节点由空值分隔（请参见示例）。示例 1:输入：root = [1,null,3,2,4,null,5,6]输出：3
题眼：二叉树的层序遍历
思路：一般 层序遍历的基础上，遍历children部分
'''class Node:def __init__(self, val=None, children=None):self.val = valself.children = childrenclass Solution:def maxDepth(self, root: Optional[Node]) -> int:# 情况1、树为空if root == None:return 0# 情况2、树不为空result = 0que = deque()que.append(root)while len(que) > 0:size = len(que)for _ in range(size):node = que.popleft()if node.children != None:for child in node.children:que.append(child)result += 1return result