binaryTree.inOrderTraversal(); System.out.println(""); binaryTree.postOrderTraversal(); }publicstaticclassBinaryTree {privateNode root;publicvoidpreOrderTraversal() {this.root.preOrderTraversal(); }publicvoidinOrderTraversal() {this.root.inOrderTraversal(); }publicvoidpostOrderTraversal() {this.ro...
* TreeNode(int x) { val = x; } * } */classSolution{public List<Integer>inorderTraversal(TreeNode root){List<Integer>result=newLinkedList<>();TreeNode current=root;TreeNode prev=null;while(current!=null){// left firstif(current.left==null){result.add(current.val);current=current.right...
* @param root: A Tree * @return: Inorder in ArrayList which contains node values. */vector<int>inorderTraversal(TreeNode * root){// write your code herevector<int> result; traverse(root, result);returnresult; }// 遍历法的递归函数,没有返回值,函数参数result贯穿整个递归过程用来记录遍历的结...
子树递归同理 中序Inorder: 先访问左子树,然后访问根节点,最后访问右子树. 后序Postorder:先访问左子树,然后访问右子树,最后访问根节点. classNode:def__init__(self,key):self.left=Noneself.right=Noneself.val=keydefprintInorder(root):ifroot:printInorder(root.left)print(root.val)printInorder(root.r...
Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level). Example 1: Input: root = [3,9,20,null,null,15,7] Output: [[3],[9,20],[15,7]] Example 2: Input: root = [1] Output: [[1]] Example ...
LeetCode之“树”:Binary Tree Preorder && Inorder && Postorder Traversal,BinaryTreePreorderTraversal题目链接题目要求:Givenabinarytree,returnthepreordertraversalofitsnodes'values.Forexample:Givenbinarytree{1...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution{public:vector<int>inorderTraversal(TreeNode*root){vector<int>res;inorder(root,res);return...
voidTree::n_preorder(){ Node* node; stack nodeStack; cout <<"\nPreorder:"; nodeStack.push(root);while(!nodeStack.isEmpty()) { node = nodeStack.pop(); cout <<" "<< node->data;// As a stack is LIFO (last-in-first-out), we add the node's children// on the stack in re...
18 A and B are two nodes on a binary tree. In the in-order traversal, the condition for A before B is ( ). A. A is to the left of B B. A is to the right of B C. A is the ancestor of B D. A is a descendant of B ...
In this article we will focus on the binary tree traversal using depth first search. 2. Depth-First-Search The depth-first search (DFS) is a tree traversal technique. In DFS, we go as deep as possible down to one path before we explore or visit the different node or the next sibling...