LeetCode二叉树实现

Posted 2020-11-20 NSGUF

LeetCode二叉树实现

# 定义二叉树
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None 

树的遍历介绍

1. 前序遍历

   前序遍历首先访问根节点，然后遍历左子树，最后遍历右子树。

2. 中序遍历

   中序遍历是先遍历左子树，然后访问根节点，然后遍历右子树。

3. 后续遍历

   后序遍历是先遍历左子树，然后遍历右子树，最后访问树的根节点。

4. 层次遍历

   该算法从一个根节点开始，首先访问节点本身。 然后遍历它的相邻节点，其次遍历它的二级邻节点、三级邻节点，以此类推。

   class Solution:
   # 前序遍历
   def solvePre(self, root, result):
   if root == None:
   return []
   result.append(int(root.val)) # 先添加根节点
   self.solvePre(root.left, result) # 再添加左子树
   self.solvePre(root.right, result) # 再添加又子树
   return result

   def preorderTraversal(self, root):
       """
       :type root: TreeNode
       :rtype: List[int]
       """
       return self.solvePre(root, [])
   
   # 中序遍历
   def solveIno(self, root, result):
       if root == None:
           return []
       self.solveIno(root.left, result)  # 先遍历左子树
       result.append(int(root.val))
       self.solveIno(root.right, result)  # 再遍历右子树
       return result
   
   def inorderTraversal(self, root):
       """
       :type root: TreeNode
       :rtype: List[int]
       """
       return self.solveIno(root, [])
   
   # 后序遍历
   def solvePos(self, root, result):
       if root == None:
           return []
       self.solvePos(root.left, result)  # 先访问左子树
       self.solvePos(root.right, result)  # 在访问右子树
       result.append(int(root.val))
       return result
   
   def postorderTraversal(self, root):
       """
       :type root: TreeNode
       :rtype: List[int]
       """
       return self.solvePos(root, [])
   
   # 层次遍历
   def levelOrder(self, root):
       """
       :type root: TreeNode
       :rtype: List[List[int]]
       """
       if root == None:
           return []
       queue = [root]  # 通过list存储
       result = []
       index = 0
       while True:
           start = index  # 该层的节点在list中的开始位置
           end = len(queue)  # 该层的节点在list中的最后位置
           block = []  # 存储该层的数据
           for i in range(start, end):
               block.append(queue[i].val)
               if queue[i].left != None:
                   queue.append(queue[i].left)
               if queue[i].right != None:
                   queue.append(queue[i].right)
               index += 1
           if start >= end:  # 如果list中的元素被循环完了，即没有添加新的节点
               break
           result.append(block)
       return result
   
   # 最大深度
   def maxDepth(self, root):
       """
       :type root: TreeNode
       :rtype: int
       """
       if root == None:
           return 0
       leftCount = 1  # 左子树的深度
       rightCount = 1  # 右子树的深度
       leftCount = leftCount + self.maxDepth(root.left)
       rightCount = rightCount + self.maxDepth(root.right)
       return max(leftCount, rightCount)  # 选深度大的
   
   # 对称二叉树
   def isSymmetric(self, root):
       """
       :type root: TreeNode
       :rtype: bool
       """
       if root == None:
           return True
       index = 0
       queue = [root]
       while True:  # 获取每层的数据存入list，如果list翻转后和list不一样 则表示不是对称的
           start = index
           end = len(queue)
           block = []
           for i in range(start, end):
               if queue[i] == None:
                   block.append(‘ ‘)  # 这是为了确定某个地方为空的，里面的数据只有不为数字就行
               else:
                   block.append(queue[i].val)
                   queue.append(queue[i].left)
                   queue.append(queue[i].right)
               index += 1
   
           if block[::-1] != block:
               return False
           if index >= len(queue):
               break
           if block.count(‘ ‘) == len(block):  # 当该层的数据都是空的，则跳出
               break
       return True
   
   def solveHasPathSum(self, root, sumAll, sum, flag):  # 使用深度搜索
       if root:
           sumAll += root.val
           if root.left or root.right:  # 当该节点不是叶子节点时
               flag = self.solveHasPathSum(root.left, sumAll, sum, flag) | self.solveHasPathSum(root.right, sumAll,
                                                                                                sum, flag)
           else:
               if sumAll == sum:
                   return True
       return flag
   
   # 路径总和
   def hasPathSum(self, root, sum):
       """
       :type root: TreeNode
       :type sum: int
       :rtype: bool
       """
       return self.solveHasPathSum(root, 0, sum, False)
   
   # 100. 相同的树
   def isSameTree(self, p, q):
       """
       :type p: TreeNode
       :type q: TreeNode
       :rtype: bool
       """
       if p == None and q == None:
           return True
       if p != None and q != None and p.val == q.val:  # 如果pq相同，则判断他们的子节点是否也相同
           return self.isSameTree(p.left, q.left) & self.isSameTree(p.right, q.right)
       return False
   
   # 从前序与中序遍历序列构造二叉树
   def buildTree(self, preorder, inorder):
       """
       :type preorder: List[int]
       :type inorder: List[int]
       :rtype: TreeNode
       """
       if preorder == []:
           return None
       root = TreeNode(preorder[0])  # 前序遍历的第一个节点是父节点
       index = inorder.index(preorder[0])  # 父节点的左边是左子树的，右边是右子树的
       root.left = self.buildTree(preorder[1:index + 1], inorder[:index])
       root.right = self.buildTree(preorder[index + 1:], inorder[index + 1:])
       return root
   
   # 从中序与后序遍历序列构造二叉树
   def buildTree(self, inorder, postorder):
       """
       :type preorder: List[int]
       :type inorder: List[int]
       :rtype: TreeNode
       """
       if postorder == []:
           return None
       root = TreeNode(postorder[-1])  # 后序遍历的最后一个节点是父节点
       index = inorder.index(postorder[-1])  # 父节点的左边是左子树的，右边是右子树的
       root.left = self.buildTree(inorder[:index], postorder[:index])
       root.right = self.buildTree(inorder[index + 1:], postorder[index:-1])
       return root
   
   # 每个节点的右向指针 II
   def connect(self, root):
       if root != None:
           index = 0
           queue = [root]
           while True:
               start = index
               end = len(queue)
               for i in range(start, end):
                   if i != end - 1:  # 如果该层的最后一个节点，则指向空，否则指向后一个节点
                       queue[i].next = queue[i + 1]
                   else:
                       queue[i].next = None
                   if queue[i].left:  # 节点存在则添加，否则不添加
                       queue.append(queue[i].left)
                   if queue[i].right:
                       queue.append(queue[i].right)
                   index += 1
               if index >= len(queue):
                   break
   
   def isContent(self, root, num):  # 判断该树是否有该节点
       if root == None:
           return False
       if root.val == num.val:
           return True
       return self.isContent(root.left, num) | self.isContent(root.right, num)
   
   # 二叉树的最近公共祖先
   def lowestCommonAncestor(self, root, p, q):
       """
       :type root: TreeNode
       :type p: TreeNode
       :type q: TreeNode
       :rtype: TreeNode
       """
       if root == None or root.val == p.val or root.val == q.val:  # 如果根节点是p或q则该节点是lca
           return root
       left = self.lowestCommonAncestor(root.left, p, q)
       right = self.lowestCommonAncestor(root.right, p, q)
       if left and right:
           return root
       return left if left != None else right
   
   # 序列化
   def serialize(self, root):
       """Encodes a tree to a single string.
       :type root: TreeNode
       :rtype: str
       """
       if root == None:
           return []
       queue = [root]
       result = []
       index = 0
       nullIndex = 0
       while True:
           start = index
           end = len(queue)
           flag = True
           for i in range(start, end):
               if queue[i] != None:
                   flag = False
                   continue
           if flag == False:
               for i in range(start, end):
                   if queue[i] == None:
                       result.append(‘null‘)
                   else:
                       result.append(queue[i].val)
                       nullIndex = i
                       if queue[i].left != None:
                           queue.append(queue[i].left)
                       else:
                           queue.append(None)
                       if queue[i].right != None:
                           queue.append(queue[i].right)
                       else:
                           queue.append(None)
                   index += 1
           else:
               break
   
       return result[:nullIndex + 1]
   
   # 反序列化
   def deserialize(self, data):
       """Decodes your encoded data to tree.
       :type data: str
       :rtype: TreeNode
       """
       inputValues = data
       if data == []:
           return None
       root = TreeNode((inputValues[0]))
       nodeQueue = [root]
       front = 0
       index = 1
       while index < len(inputValues):
           node = nodeQueue[front]
           front = front + 1
   
           item = inputValues[index]
           index = index + 1
           if item != "null":
               leftNumber = int(item)
               node.left = TreeNode(leftNumber)
               nodeQueue.append(node.left)
   
           if index >= len(inputValues):
               break
   
           item = inputValues[index]
           index = index + 1
           if item != "null":
               rightNumber = int(item)
               node.right = TreeNode(rightNumber)
               nodeQueue.append(node.right)
       return root