代码随想录算法训练营【二叉树篇层序遍历】

二叉树的层序遍历

注：本文代码来自于代码随想录

102.二叉树的层序遍历

力扣102

Python

长度法

# 利用长度法
# 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
class Solution:def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:if not root:return []queue = collections.deque([root])result = []while queue:level = []for _ in range(len(queue)):cur = queue.popleft()level.append(cur.val)if cur.left:queue.append(cur.left)if cur.right:queue.append(cur.right)result.append(level)return result

递归法

#递归法
# 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
class Solution:def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:if not root:return []levels = []def traverse(node, level):if not node:returnif len(levels) == level:levels.append([])levels[level].append(node.val)traverse(node.left, level + 1)traverse(node.right, level + 1)traverse(root, 0)return levels

107.二叉树的层次遍历 II

力扣107

Python

# 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
class Solution:def levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:if not root:return []queue = collections.deque([root])result = []while queue:level = []for _ in range(len(queue)):cur = queue.popleft()level.append(cur.val)if cur.left:queue.append(cur.left)if cur.right:queue.append(cur.right)result.append(level)return result[::-1]

199.二叉树的右视图

力扣199

层序遍历的时候，判断是否遍历到单层的最后面的元素，如果是，就放进result数组中，随后返回result就可以了。

Python

# 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
class Solution:def rightSideView(self, root: TreeNode) -> List[int]:if not root:return []queue = collections.deque([root])right_view = []while queue:level_size = len(queue)for i in range(level_size):node = queue.popleft()if i == level_size - 1:right_view.append(node.val)if node.left:queue.append(node.left)if node.right:queue.append(node.right)return right_view

637.二叉树的层平均值

力扣637

Python

class Solution:"""二叉树层平均值迭代解法"""# 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
class Solution:def averageOfLevels(self, root: TreeNode) -> List[float]:if not root:return []queue = collections.deque([root])averages = []while queue:size = len(queue)level_sum = 0for i in range(size):node = queue.popleft()level_sum += node.valif node.left:queue.append(node.left)if node.right:queue.append(node.right)averages.append(level_sum / size)return averages

429.N叉树的层序遍历

力扣429

Python

层序遍历

"""
# Definition for a Node.
class Node:def __init__(self, val=None, children=None):self.val = valself.children = children
"""class Solution:def levelOrder(self, root: 'Node') -> List[List[int]]:if not root:return []result = []queue = collections.deque([root])while queue:level_size = len(queue)level = []for _ in range(level_size):node = queue.popleft()level.append(node.val)for child in node.children:queue.append(child)result.append(level)return result

递归法

# LeetCode 429. N-ary Tree Level Order Traversal
# 递归法
class Solution:def levelOrder(self, root: 'Node') -> List[List[int]]:if not root: return []result=[]def traversal(root,depth):if len(result)==depth:result.append([])result[depth].append(root.val)if root.children:for i in range(len(root.children)):traversal(root.children[i],depth+1)traversal(root,0)return result

515.在每个树行中找最大值

力扣515

Python

# 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
class Solution:def largestValues(self, root: TreeNode) -> List[int]:if not root:return []result = []queue = collections.deque([root])while queue:level_size = len(queue)max_val = float('-inf')for _ in range(level_size):node = queue.popleft()max_val = max(max_val, node.val)if node.left:queue.append(node.left)if node.right:queue.append(node.right)result.append(max_val)return result

116.填充每个节点的下一个右侧节点指针

力扣116

本题依然是层序遍历，只不过在单层遍历的时候记录一下本层的头部节点，然后在遍历的时候让前一个节点指向本节点就可以了

Python

"""
# Definition for a Node.
class Node:def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):self.val = valself.left = leftself.right = rightself.next = next
"""
class Solution:def connect(self, root: 'Node') -> 'Node':if not root:return rootqueue = collections.deque([root])while queue:level_size = len(queue)prev = Nonefor i in range(level_size):node = queue.popleft()if prev:prev.next = nodeprev = nodeif node.left:queue.append(node.left)if node.right:queue.append(node.right)return root

117.填充每个节点的下一个右侧节点指针II

力扣117

这道题目说是二叉树，但116题目说是完整二叉树，其实没有任何差别，一样的代码一样的逻辑一样的味道

Python

# 层序遍历解法
"""
# Definition for a Node.
class Node:def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):self.val = valself.left = leftself.right = rightself.next = next
"""class Solution:def connect(self, root: 'Node') -> 'Node':if not root:return rootqueue = collections.deque([root])while queue:level_size = len(queue)prev = Nonefor i in range(level_size):node = queue.popleft()if prev:prev.next = nodeprev = nodeif node.left:queue.append(node.left)if node.right:queue.append(node.right)return root

104.二叉树的最大深度

力扣104

Python

# 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
class Solution:def maxDepth(self, root: TreeNode) -> int:if not root:return 0depth = 0queue = collections.deque([root])while queue:depth += 1for _ in range(len(queue)):node = queue.popleft()if node.left:queue.append(node.left)if node.right:queue.append(node.right)return depth

111.二叉树的最小深度

力扣111

Python

# 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
class Solution:def minDepth(self, root: TreeNode) -> int:if not root:return 0depth = 0queue = collections.deque([root])while queue:depth += 1 for _ in range(len(queue)):node = queue.popleft()if not node.left and not node.right:return depthif node.left:queue.append(node.left)if node.right:queue.append(node.right)return depth