链接:https://leetcode-cn.com/problems/binary-tree-inorder-traversal python # 0094.二叉树中序遍历 # 递归 & 迭代 class Solution: def inOrderRecur(self,head: TreeNode) -> int: """ 递归遍历,LNR, 左根右 :param head: :return: """ def traversal(head): # 递归终止条件 ifhead== None: ...
中序遍历的顺序是:先遍历左子树,然后访问根节点,最后遍历右子树。这种遍历方式得到的BST中的元素是按照升序排列的。 以下是一个使用递归实现中序遍历的Python示例代码: python def inorderTraversal(root): if root is None: return inorderTraversal(root.left) # 遍历左子树 print(root.val) # 访问根节点 ino...
代码: #Definition for a binary tree node#class TreeNode:#def __init__(self, x):#self.val = x#self.left = None#self.right = NoneclassSolution:#@param root, a tree node#@return a list of integersdefiterative_inorder(self, root, list): stack=[]whilerootorstack:ifroot: stack.append...
T = Tree()foriinarr: T.insert(i)print('BST in-order traversal---') T.traversal()print('\ndelete test---')foriinarr[::-1]:print('after delete',i,end=',BST in-order is = ') T.delete(i) T.traversal()print() 输出结果为: BSTin-order traversal---0123456789deletetest---afterd...
self.inorderTraversal(root.right) def isValidBST(self, root): """ :type root: TreeNode :rtype: bool """ if not root: return True # if not root.left or not root.right: # return True self.inorderTraversal(root) return self.flag ...
中序遍历 InOrder Traversal：左结点-根-右结点后序遍历 PostOrder Traversal：左结点-右结点-根 广度优先搜索（Breadth First Search, BFS）层次遍历 LevelOrder Traversal def level_order(root): if not root: return [] res = [] nodequeue = [root] while nodequeue: root = nodequeue.pop...
Write a Python program to delete a node with a given key from a BST and then perform an in-order traversal to verify the tree remains valid. Write a Python script to remove a node from a BST, handling all three cases (leaf, one child, two children), and then print the ...
Python Java C C++ # Binary Search Tree operations in Python# Create a nodeclassNode:def__init__(self, key):self.key = key self.left =Noneself.right =None# Inorder traversaldefinorder(root):ifrootisnotNone:# Traverse leftinorder(root.left)# Traverse rootprint(str(root.key) +"->", ...
094.binary-tree-inorder-traversal Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree [1,null,2,3], 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively?
The code below is an implementation of the Binary Search Tree in the figure above, with traversal.Example Python: class TreeNode: def __init__(self, data): self.data = data self.left = None self.right = None def inOrderTraversal(node): if node is None: return inOrderTraversal(node....