每天刷个算法题20160521:二叉树高度(递归与非递归)

Posted 小飞_Xiaofei

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了每天刷个算法题20160521:二叉树高度(递归与非递归)相关的知识,希望对你有一定的参考价值。

版权所有。所有权利保留。

欢迎转载,转载时请注明出处:

http://blog.csdn.net/xiaofei_it/article/details/51502727


为了防止思维僵化,每天刷个算法题。已经刷了几天了,现在发点代码。

我已经建了一个开源项目,每天的题目都在里面:

https://github.com/Xiaofei-it/Algorithms

绝大部分算法都是我自己写的,没有参考网上通用代码。读者可能会觉得有的代码晦涩难懂,因为那是我自己的理解。

最近几天都是在写一些原来的东西,大多数是非递归。以后准备刷点DP、贪心之类的题。

下面是二叉树高度(递归与非递归):

/**
 *
 * Copyright 2016 Xiaofei
 *
 * Licensed under the Apache License, Version 2.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 * http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 *
 */

package xiaofei.algorithm;

import java.util.Stack;

/**
 * Created by Xiaofei on 16/5/21.
 *
 * I am writing code in Starbucks alone.
 *
 * What a sad picture!!!
 *
 * I cannot remember when I drank coffee with somebody else last time.
 *
 */
public class BinaryTreeHeight 

    public static int calculateRecursively(BinaryTreeNode root) 
        if (root == null) 
            return 0;
         else 
            return Math.max(calculateRecursively(root.left), calculateRecursively(root.right)) + 1;
        
    

    public static int calculateCorecursively(BinaryTreeNode root) 
        //当年ACM的时候总喜欢自己写queue或者stack,几乎不用C++的STL库。现在想用一下Java的Stack
        class Element 
            BinaryTreeNode node;
            int left;
            int right;
            Element(BinaryTreeNode node) 
                this.node = node;
                left = right = -1;
            
        
        Stack<Element> stack = new Stack<>();
        stack.push(new Element(root));
        int result = -1;
        while (!stack.empty()) 
            Element element = stack.peek();
            if (element.node == null) 
                stack.pop();
                if (stack.isEmpty()) 
                    result = 0;
                 else 
                    Element tmp = stack.peek();
                    if (tmp.left == -1) 
                        tmp.left = 0;
                     else 
                        tmp.right = 0;
                    
                
             else if (element.left == -1) 
                stack.push(new Element(element.node.left));
             else if (element.right == -1) 
                stack.push(new Element(element.node.right));
             else 
                Element tmp1 = stack.pop();
                int height = Math.max(tmp1.left, tmp1.right) + 1;
                if (stack.isEmpty()) 
                    result = height;
                 else 
                    Element tmp2 = stack.peek();
                    if (tmp2.left == -1) 
                        tmp2.left = height;
                     else 
                        tmp2.right = height;
                    
                
            
        
        return result;
    



以上是关于每天刷个算法题20160521:二叉树高度(递归与非递归)的主要内容,如果未能解决你的问题,请参考以下文章

每天刷个算法题20160521:二叉树高度(递归与非递归)

每天刷个算法题20160518:非递归二叉树遍历

每天刷个算法题20160518:非递归二叉树遍历

每天刷个算法题20160520:二叉排序树

每天刷个算法题20160520:二叉排序树

每天刷个算法题20160520:二叉排序树