self.inorder(root.right, list)definorderTraversal(self, root): list=[] self.iterative_inorder(root, list)returnlist
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: ...
importsysimportioclassSolution:def__init__(self):self.min_node=sys.maxsize+1self.res_path=[]self.tree=[]defsmallest_node(self,input_values:list)->list:self.tree=input_values# Find min and add -1 to all empty child nodesforiinrange(len(self.tree)):if0<self.tree[i]<self.min_node:...
代码(Python3) # 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 inorderTraversal(self, root: Optional[TreeNode]) -> List[int]: # ans 用...
Python代码如下: # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def buildTree(self, preorder, inorder): ...
leetcode 94.二叉树的中序遍历(binary tree inorder traversal)C语言 1.description 2.solution 2.1 递归 2.2 迭代 1.description https://leetcode-cn.com/problems/binary-tree-inorder-traversal/ 给定一个二叉树,返回它的中序 遍历。 示例: 输入: [...猜...
However, the pre-order traversal is often used when we want to modify the tree during the traversal since it’s easy to alter the root node first and then the left and right child nodes. Here’s a syntax and simple implementation of the in-order traversal in Python. ...
The in-order traversal of whole tree is : 11, 20, 22, 50, 52, 53, 78. Implementation of In-order tree traversal in Python The algorithm for in-order tree traversal can be formulated as follows. Recursively Traverse the left sub-tree. Print the root node. Recursively Traverse the right...
(Tree, element) return Tree class Solution(object): def postorderTraversal(self, root): if not root: return [] res = [] stack = [[root,0]] while stack: node = stack[-1] stack.pop() if node[1]== 0 : current = node[0] stack.append([current,1]) if current.right: stack....
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?