遍历二叉树的神级方法

【说明】：

　　本文是左程云老师所著的《程序员面试代码指南》第三章中“遍历二叉树的神级方法”这一题目的C++复现。

　　本文只包含问题描述、C++代码的实现以及简单的思路，不包含解析说明，具体的问题解析请参考原书。

　　感谢左程云老师的支持。

【题目】：

　　给定一个二叉树的头节点 head，完成二叉树的先序、中序和后序遍历。如果二叉树的节点数为N，要求时间复杂度为O(N)，额外的空间复杂度为O(1)。

 【思路】：

　　解法：Morris 遍历方法，利用空节点。

【编译环境】：

　　CentOS6.7（x86_64）

　　gcc 4.4.7

 【实现及测试】：

　　声明代码：

/*
 *Filename:bt_morris.h
 *Author:
 *Summary:Use Morris method to traversing binary tree.
 */

#include <iostream>
using namespace std;

class Node
{
public:
    Node(int data)
    {
        value = data;
        left = NULL;
        right = NULL;
    }
public:
    int value;
    Node *left;
    Node *right;
};

void morrisIn(Node *head);  // inorder traversal
void morrisPre(Node *head);    // preorder traversal 
void morrisPos(Node *head); // posorder traversal

　　实现及测试代码：

/*
 *Filename:bt_morris.h
 *Author:
 *Summary:Use Morris method to traversing binary tree.
 */

#include "bt_morris.h"


void morrisIn(Node *head)
{
    if (NULL == head)
        return;

    Node *cur1 = head;
    Node *cur2 = NULL;
    while (NULL != cur1)
    {
        cur2 = cur1->left; // cur2 is cur1's left node.
        if (NULL != cur2)
        {
            while(NULL != cur2->right && cur2->right != cur1)  //Find rightmost node of cur2 and detect if this node point to cur1.
            {
                cur2 = cur2->right;
            }
            if (NULL == cur2->right)    
            {
                cur2->right = cur1;   // Make the rightmost node's right-node(NULL) point to cur2.
                cur1 = cur1->left;    // Continue until cur1's left node is NULL.
                continue;
            }
            else
            {
                cur2->right = NULL;
            }
        }
        cout << cur1->value << " ";   //Print node cur1.
        cur1 = cur1->right;   //Move to the cur1's right node.
    }
    cout << endl;
}


void morrisPre(Node *head)
{
    if (NULL == head)
        return;

    Node *cur1 = head;
    Node *cur2 = NULL;
    while (NULL != cur1)
    {
        cur2 = cur1->left; // cur2 is cur1's left node.
        if (NULL != cur2)
        {
            while(NULL != cur2->right && cur2->right != cur1)  //Find rightmost node of cur2 and detect if this node point to cur1.
            {
                cur2 = cur2->right;
            }
            if (NULL == cur2->right)    
            {
                cur2->right = cur1;   // Make the rightmost node's right-node(NULL) point to cur2.
                cout << cur1->value << " ";   //Print node cur1.
                cur1 = cur1->left;    // Continue until cur1's left node is NULL.
                continue;
            }
            else
            {
                cur2->right = NULL;
            }
        }
        else
        {
            cout << cur1->value << " ";
        }
        cur1 = cur1->right;   //Move to the cur1's right node.
    }
    cout << endl;
}


Node* reverseEdge(Node *from)   //Since node from, reverse the right border.
{
    Node *pre = NULL;
    Node *next = NULL;
    while(NULL != from)
    {
        next = from->right;
        from->right = pre;
        pre = from;
        from = next;
    }
    return pre;      //In the end, node pre (the rightmost node), become the head of the border.

}

void printEdge(Node *head)
{
    Node *tail = reverseEdge(head);
    Node *cur = tail;
    while(NULL != cur)
    {
        cout << cur->value << " ";
        cur =  cur->right;    
    }
    reverseEdge(tail);  //reverse and reverse, the border change back to the original.
    return;
}


void morrisPos(Node *head)
{
    if (NULL == head)
        return;

    Node *cur1 = head;
    Node *cur2 = NULL;
    while (NULL != cur1)
    {
        cur2 = cur1->left; // cur2 is cur1's left node.
        if (NULL != cur2)
        {
            while(NULL != cur2->right && cur2->right != cur1)  //Find rightmost node of cur2 and detect if this node point to cur1.
            {
                cur2 = cur2->right;
            }
            if (NULL == cur2->right)    
            {
                cur2->right = cur1;   // Make the rightmost node's right-node(NULL) point to cur2.
                cur1 = cur1->left;    // Continue until cur1's left node is NULL.
                continue;
            }
            else
            {
                cur2->right = NULL;
                printEdge(cur1->left);
            }
        }
        cur1 = cur1->right;   //Move to the cur1's right node.
    }
    printEdge(head);
    cout << endl;
}

void createBT(Node **head,int arr[],int len,int index=0)
{
    if(index > len-1 || -1 == arr[index] )
        return;
    (*head) = new Node(arr[index]);
    createBT(&((*head)->left),arr,len,2*index+1);
    createBT(&((*head)->right),arr,len,2*index+2);
}

int main()
{
    int arr[] = {1,2,3,4,5,6,7};
    Node *root = NULL;
    createBT(&root,arr,7);
    morrisIn(root);
    morrisPre(root);
    morrisPos(root);

    return 0;
}

注：

　　转载请注明出处；

　　转载请注明源思路来自于左程云老师的《程序员代码面试指南》。