java创建二叉树并递归遍历二叉树前面已有讲解:http://www.cnblogs.com/lixiaolun/p/4658659.html. 在此基础上添加了非递归中序遍历二叉树: 二叉树类的代码: package binarytree; import linkedstack.LinkStack; import linkqueue.LinkQueue; public class BinaryTree { class Node { public Object data; public Node
Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively?中序遍历二叉树,递归遍历当然很容易,题目还要求不用递归,下面给出两种方法: 递归: /**
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(递归) solution is trivial, could you do it iteratively(迭代)? 思路: 解法一:用递归方法很简单, (1)如果root为空,则返回
给定一个二叉树,返回其中序遍历.例如:给定二叉树 [1,null,2,3], 1 \ 2 / 3返回 [1,3,2].说明: 递归算法很简单,你可以通过迭代算法完成吗?详见:https://leetcode.com/problems/binary-tree-inorder-traversal/description/ Java实现: 递归实现: /** * Definition for a binary tree node. * public class TreeNo
Level: Medium 题目描述: Given a binary tree, return the inorder traversal of its nodes' values. 思路分析: 实现一棵二叉树的中序遍历,我们可以用简单的递归方法去实现,也可以使用栈去实现,使用第二种方式时,我们沿着根节点先遍历左子树的左孩子,将它们依次压入栈,知道左孩子为空,弹出栈顶节点,这时记录栈顶节点的值,如果栈顶节点的右孩子不为空,压入栈,如果为空,则栈顶元素继续弹出,重复上述操作,就能获得中序遍
/** * 实现二叉树的创建.前序遍历.中序遍历和后序遍历 **/ package DataStructure; /** * Copyright 2014 by Ruiqin Sun * All right reserved * created on 2014-9-9 下午2:34:15 **/ public class BinTreeInt { private Node root; /** * 创建内部节点类 **/ private class Node{ // 左节点 private Nod
package com.example.demo; public class BTree { public int data; public BTree left; public BTree rigth; public boolean hasLeft(){ return left != null; } public boolean hasRigth(){ return rigth != null; } public BTree(){} } class main{ public static vo
Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 这道题要求用先序和中序遍历来建立二叉树,跟之前那道Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树原理基本相同,针对这道题,由于先
Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? confused what "{1,#,2,3}" means? > re
