力扣_二叉樹卡片題目答案彙總
阿新 • • 發佈：2020-08-05
題目來源
這裡只是附上各個程式碼答案，並不做解釋，至少現在不做解釋。
程式碼
二叉樹類
from collections import deque # 數的節點類 class TreeNode(object): def __init__(self, item): self.val = item self.left = None self.right = None # python 二叉樹類 class Tree(object): def __init__(self): self.root = None self.res = 0 def construct_tree(self, values=None): if not values: return None self.root = TreeNode(values[0]) queue = deque([self.root]) leng = len(values) nums = 1 while nums < leng: node = queue.popleft() if node: node.left = TreeNode(values[nums]) if values[nums] else None queue.append(node.left) if nums + 1 < leng: node.right = TreeNode(values[nums + 1]) if values[nums + 1] else None queue.append(node.right) nums += 1 nums += 1 def bfs(self): res = [] queue = deque([self.root]) while queue: node = quque.popleft() if node: res.append(node.val) queue.append(node.left) queue.append(node.right) return ret # 迭代遍歷，資料儲存一層一個列表 # if not root: # return [] # queue = deque([self.root]) # ret = [] # while queue: # list_1 = [] # for i in range(len(queue)): # node = queue.popleft() # if node.left: # queue.append(node.left) # if node.right: # queue.append(node.right) # list_1.append(node.val) # ret.append(list_1) # return ret def pre_traversal(self): ret = [] # 遞迴方式 # def traversal(head): # if not head: # return # ret.append(head.val) # traversal(head.left) # traversal(head.right) # traversal(self.root) # 迭代方式 if not root: return [] stack = [self.root] while stack: node = stack.pop() ret.append(node.val) if node.right: # 先加右子節點，因為列表pop先拋卻最右的， stack.append(node.right) if node.left: stack.append(node.left) return ret def in_traversal(self): # 遞迴方式 # ret = [] # def traversal(head): # if not head: # return # traversal(head.left) # ret.append(head.val) # traversal(head.right) # traversal(self.root) # 迭代方式 if not root: return [] ret = [] stack = [] root = self.root while stack or root: while root: stack.append(root) root = root.left root = stack.pop() ret.append(root.val) root = root.right return ret def pre_traversal(self): # 遞迴方式 # ret = [] # def traversal(head): # if not head: # return # traversal(head.left) # traversal(head.right) # ret.append(head.val) # traversal(self.root) # return ret # 迭代方式 if not root: return [] stack = [self.root] ret = [] while stack: node = stack.pop() ret.append(node.val) if node.left: stack.append(node.left) if node.right: stack.append(node.right) return result[::-1] def get_depth(self, node): # 自底向上 if not node: return 0 return max(self.get_depth(node.left) + 1, self.get_depth(node.right) + 1) # 自頂向下 def get_depth2(self, node, depth): if not node: return print(f"res===>{self.res}, depth===>{depth}") if not node.left and not node.right: self.res = max(self.res, depth) depth += 1 self.get_depth2(node.left, depth) self.get_depth2(node.right, depth) return self.res
問題1 二叉樹的最大深度
class Solution:
def maxDepth(self, root):
# 方法一：自底向上
if not root:
return 0
return max(self.maxDepth(root.left) + 1, self.maxDepth(root.right) + 1)
問題2 對稱二叉樹
from collections import deque class Solution: def isSymmetric(self, root): # 方法一：遞迴 if not root: return True return self.isST(root.left, root.right) def isST(self, left, right): if not left and not right: return True if not left or not right or left.val != right.val: return False return self.isST(left.left, right.right) and self.isST(left.right, right.left)
問題3 路徑總和
class Solution: def hasPathSum(self, root, Pathsum): # 方法一：DFS if not root: return False if not root.left and not root.right: return sum == root.val return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)
問題4 從中序遍歷和後序遍歷構造二叉樹
class Solution:
def buildTree(self, inorder, postorder):
if not inorder or not postorder:
return None
root = TreeNode(postorder[-1]) # 根節點
index = inorder.index(postorder[-1])
root.left = self.buildTree(inorder[:index], postorder[:index])
root.right = self.buildTree(inorder[index + 1:], postorder[index:-1])
return root
問題5 從前序與中序遍歷序列構造二叉樹
class Solution:
def buildTree(self, preorder, inorder):
if not preorder or not inorder:
return None
root = TreeNode(preorder[0])
index = inorder.index(preorder[0])
root.left = self.buildTree(preorder[1: 1 + index], inorder[:index])
root.right = self.buildTree(preorder[index + 1:], inorder[index+1:])
return root
問題6 填充每個節點到下一個右側節點指標
class Solution:
def connect(self, root): # root: 'Node' -> 'Node'
# 遞迴版：
if not root or not root.left:
return root
root.left.next = root.right
if root.next and root.right:
root.right.next = root.next.left
self.connect(root.left)
self.connect(root.right)
return root
# 迭代版
if not root:
return
the_root = root
while root.left is not None:
next_layer = root.left
while root.next is not None:
root.left.next = root.right
root.right.next = root.next.left
root = root.next
root.left.next = root.right
root = next_layer
return the_root
問題7 填充每個節點的下一個右側節點指標II
class Solution:
def connect(self, root): # root: 'Node' -> 'Node'
# 迭代版
head = root
while head:
cur = head
pre = head = None
while cur:
if cur.left:
if not pre:
pre = head = cur.left
else:
pre.next = cur.left
pre = pre.next
if cur.right:
if not pre:
pre = head = cur.right
else:
pre.next = cur.right
pre = pre.next
cur = cur.next
return root
問題8 二叉樹的最近公共祖先
class Solution:
def lowestCommonAncestor(self, root, p, q): # root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:
if root is None:
return None
if root == p or root == q:
return root
left = self.lowestCommonAncestor(root.left, p, q)
right = self.lowestCommonAncestor(root.right, p, q)
if left and right: # p q 分別在左右兩子樹中
return root
if not left: # p q 都在右子樹中
return right
if not right: # p q 都在左子樹中
return left
問題9 二叉樹的序列化與反序列化
from collections import deque
class Codec:
def serialize(self, root):
"""Encodes a tree to a single string.
:type root: TreeNode
:rtype: str
"""
s = ""
stack = [root]
while stack:
root = stack.pop(0)
if root:
s += str(root.val)
stack.append(root.left)
stack.append(root.right)
else:
s += "n"
s += " "
return s
def deserialize(self, data):
"""Decodes your encoded data to tree.
:type data: str
:rtype: TreeNode
"""
tree = data.split()
if tree[0] == "n":
return None
stack = []
root = TreeNode(int(tree[0]))
stack.append(root)
i = 1
while stack:
cur = stack.pop(0)
if cur == None:
continue
cur.left = TreeNode(int(tree[i])) if tree[i] != "n" else None
cur.right = TreeNode(int(tree[i + 1])) if tree[i + 1] != "n" else None
i += 2
stack.append(cur.left)
stack.append(cur.right)
return root