【数据结构】70_二叉树的经典面试题分析

头像
TianSong
    阅读 11 分钟

    面试题 一

    单结点删除
    描述：编写一个函数用于删除二叉树中的所有单度结点
    要求：结点删除后，其唯一的子结点替代它的位置

    image.png

    情形1：结点中包含指向父结点的指针

    • 定义功能：delOdd1(node)

      • 删除 node 为根结点的二叉树中的单度结点

    image.png

    编程实验：单度结点的删除

    #include <iostream>
#include "BTreeNode.h"

using namespace std;
using namespace DTLib;

template < typename T >
BTreeNode<T>* createTree()
{
    static BTreeNode<int> ns[9];

    for(int i=0; i<9; i++)
    {
        ns[i].value = i;
        ns[i].parent = NULL;
        ns[i].left = NULL;
        ns[i].right = NULL;
    }

    ns[0].left = &ns[1];
    ns[0].right = &ns[2];
    ns[1].parent = &ns[0];
    ns[2].parent = &ns[0];

    ns[1].left = &ns[3];
    ns[1].right = NULL;
    ns[3].parent = &ns[1];

    ns[2].left = &ns[4];
    ns[2].right = &ns[5];
    ns[4].parent = &ns[2];
    ns[5].parent = &ns[2];

    ns[3].left = NULL;
    ns[3].right = &ns[6];
    ns[6].parent = &ns[3];

    ns[4].left = &ns[7];
    ns[4].right = NULL;
    ns[7].parent = &ns[4];

    ns[5].left = &ns[8];
    ns[5].right = NULL;
    ns[8].parent = &ns[5];

    return ns;
}

template < typename T >
void printInOrder(BTreeNode<T>* node)
{
    if( node != NULL )
    {
        printInOrder(node->left);

        cout << node->value <<" ";

        printInOrder(node->right);
    }
}

template < typename T >
void printDualList(BTreeNode<T>* node)
{
    BTreeNode<T>* g = node;

    cout << "head -> tail: " << endl;

    while( node != NULL )
    {
        cout << node->value << " ";

        g = node;

        node = node->right;
    }

    cout << endl;

    cout << "tail -> head: " << endl;

    while( g != NULL )
    {
        cout << g->value << " ";

        g = g->left;
    }

    cout << endl;
}

//----------------------------------------------------------------------
template <typename T>
BTreeNode<T>* delOdd1(BTreeNode<T> *node)
{
    BTreeNode<T> *ret = nullptr;

    if (node != nullptr)
    {
        if (((node->left != nullptr) && (node->right == nullptr)) ||
            ((node->left == nullptr) && (node->right != nullptr)))
        {
            BTreeNode<T> *parent = dynamic_cast<BTreeNode<T>*>(node->parent);
            BTreeNode<T> *child = (node->left != nullptr) ? node->left : node->right;

            if (parent != nullptr)
            {
                BTreeNode<T> *& parent_child = (parent->left == node) ? parent->left : parent->right;

                parent_child = child;
                child->parent = parent;
            }
            else
            {
                child->parent = nullptr;
            }

            if (node->flag())
            {
                delete node;
            }

            ret = delOdd1(child);
        }
        else
        {
            delOdd1(node->left);
            delOdd1(node->right);

            ret = node;
        }
    }

    return ret;
}

int main()
{
    BTreeNode<int>* ns = createTree<int>();

    printInOrder(ns);

    cout << endl;

    ns = delOdd1(ns);

    printInOrder(ns);

    return 0;
}

    输出：

    3 6 1 0 7 4 2 8 5
6 0 7 2 8

    情形2：结点中只包含左右孩子指针

    • 定义功能： delOdd2(node)

      • 删除 node 为根结点的二叉树中的单度结点

    image.png

    编程实验：单度结点删除

    #include <iostream>
#include "BTreeNode.h"

using namespace std;
using namespace DTLib;

template < typename T >
BTreeNode<T>* createTree()
{
    static BTreeNode<int> ns[9];

    for(int i=0; i<9; i++)
    {
        ns[i].value = i;
        ns[i].parent = NULL;
        ns[i].left = NULL;
        ns[i].right = NULL;
    }

    ns[0].left = &ns[1];
    ns[0].right = &ns[2];
    ns[1].parent = &ns[0];
    ns[2].parent = &ns[0];

    ns[1].left = &ns[3];
    ns[1].right = NULL;
    ns[3].parent = &ns[1];

    ns[2].left = &ns[4];
    ns[2].right = &ns[5];
    ns[4].parent = &ns[2];
    ns[5].parent = &ns[2];

    ns[3].left = NULL;
    ns[3].right = &ns[6];
    ns[6].parent = &ns[3];

    ns[4].left = &ns[7];
    ns[4].right = NULL;
    ns[7].parent = &ns[4];

    ns[5].left = &ns[8];
    ns[5].right = NULL;
    ns[8].parent = &ns[5];

    return ns;
}

template < typename T >
void printInOrder(BTreeNode<T>* node)
{
    if( node != NULL )
    {
        printInOrder(node->left);

        cout << node->value <<" ";

        printInOrder(node->right);
    }
}

template < typename T >
void printDualList(BTreeNode<T>* node)
{
    BTreeNode<T>* g = node;

    cout << "head -> tail: " << endl;

    while( node != NULL )
    {
        cout << node->value << " ";

        g = node;

        node = node->right;
    }

    cout << endl;

    cout << "tail -> head: " << endl;

    while( g != NULL )
    {
        cout << g->value << " ";

        g = g->left;
    }

    cout << endl;
}

//----------------------------------------------------------------------
template <typename T>
void delOdd2(BTreeNode<T> *&node)
{
    if (node != nullptr)
    {
        if (((node->left != nullptr) && (node->right == nullptr)) ||
            ((node->left == nullptr) && (node->right != nullptr)))
        {
            if (node->flag())
            {
                delete node;
            }

            node = (node->left != nullptr) ? node->left : node->right;

            delOdd2(node);
        }
        else
        {
            delOdd2(node->left);
            delOdd2(node->right);
        }
    }
}

int main()
{
    BTreeNode<int>* ns = createTree<int>();

    printInOrder(ns);

    cout << endl;

    delOdd2(ns);

    printInOrder(ns);

    return 0;
}

    输出：

    3 6 1 0 7 4 2 8 5
6 0 7 2 8

    面试题 二

    中序线索化二叉树
    描述：编写一个函数用于中序线索化二叉树
    要求：不允许使用其他数据结构

    image.png

    解法一：在中序遍历的同时进行线索化

    思路

    • 使用辅助指针，在中序遍历时指向当前结点的前驱结点
    • 访问当前结点时，连接与前驱结点的先后次序

    image.png

    定义功能

    • inOrderThread(node, pre)

      • node : 根结点，也是中序访问的结点
      • pre: 为中序遍历时的前驱结点指针

    image.png

    编程实验：中序线索化

    #include <iostream>
#include "BTreeNode.h"

using namespace std;
using namespace DTLib;

template < typename T >
BTreeNode<T>* createTree()
{
    static BTreeNode<int> ns[9];

    for(int i=0; i<9; i++)
    {
        ns[i].value = i;
        ns[i].parent = NULL;
        ns[i].left = NULL;
        ns[i].right = NULL;
    }

    ns[0].left = &ns[1];
    ns[0].right = &ns[2];
    ns[1].parent = &ns[0];
    ns[2].parent = &ns[0];

    ns[1].left = &ns[3];
    ns[1].right = NULL;
    ns[3].parent = &ns[1];

    ns[2].left = &ns[4];
    ns[2].right = &ns[5];
    ns[4].parent = &ns[2];
    ns[5].parent = &ns[2];

    ns[3].left = NULL;
    ns[3].right = &ns[6];
    ns[6].parent = &ns[3];

    ns[4].left = &ns[7];
    ns[4].right = NULL;
    ns[7].parent = &ns[4];

    ns[5].left = &ns[8];
    ns[5].right = NULL;
    ns[8].parent = &ns[5];

    return ns;
}

template < typename T >
void printInOrder(BTreeNode<T>* node)
{
    if( node != NULL )
    {
        printInOrder(node->left);

        cout << node->value <<" ";

        printInOrder(node->right);
    }
}

template < typename T >
void printDualList(BTreeNode<T>* node)
{
    BTreeNode<T>* g = node;

    cout << "head -> tail: " << endl;

    while( node != NULL )
    {
        cout << node->value << " ";

        g = node;

        node = node->right;
    }

    cout << endl;

    cout << "tail -> head: " << endl;

    while( g != NULL )
    {
        cout << g->value << " ";

        g = g->left;
    }

    cout << endl;
}

template <typename T>
void inOrderThread(BTreeNode<T> *node, BTreeNode<T> *&pre)
{
    if (node != nullptr)
    {
        inOrderThread(node->left, pre);

        node->left = pre;

        if (pre != nullptr)
        {
            pre->right = node;
        }

        pre = node;

        inOrderThread(node->right, pre);
    }
}

template <typename T>
BTreeNode<T> *inOrderThread(BTreeNode<T> *node)
{
    BTreeNode<T> * pre = nullptr;

    inOrderThread(node, pre);

    while ((node != nullptr) && (node->left != nullptr))
    {
        node = node->left;
    }

    return node;
}

int main()
{
    BTreeNode<int>* ns = createTree<int>();

    printInOrder(ns);

    cout << endl;

    ns = inOrderThread(ns);

    printDualList(ns);

    return 0;
}

    输出：

    3 6 1 0 7 4 2 8 5
head -> tail:
3 6 1 0 7 4 2 8 5
tail -> head:
5 8 2 4 7 0 1 6 3

    解法二：中序遍历的结点次序正好是结点的水平次序

    思路

    • 使用辅助指针，指向转换后双向链表的头结点和尾结点
    • 根结点与左右子树转换的双向链表连接，成为完整双向链表

    image.png

    定义功能

    • inOrderThread(node, head, tail)

      • node: 根结点，也是中序访问的结点
      • head: 转换成功后指向双向链表的首结点
      • tail：转换成功后指向双向链表的尾结点

    image.png

    编程实验：中序线索化

    #include <iostream>
#include "BTreeNode.h"

using namespace std;
using namespace DTLib;

template < typename T >
BTreeNode<T>* createTree()
{
    static BTreeNode<int> ns[9];

    for(int i=0; i<9; i++)
    {
        ns[i].value = i;
        ns[i].parent = NULL;
        ns[i].left = NULL;
        ns[i].right = NULL;
    }

    ns[0].left = &ns[1];
    ns[0].right = &ns[2];
    ns[1].parent = &ns[0];
    ns[2].parent = &ns[0];

    ns[1].left = &ns[3];
    ns[1].right = NULL;
    ns[3].parent = &ns[1];

    ns[2].left = &ns[4];
    ns[2].right = &ns[5];
    ns[4].parent = &ns[2];
    ns[5].parent = &ns[2];

    ns[3].left = NULL;
    ns[3].right = &ns[6];
    ns[6].parent = &ns[3];

    ns[4].left = &ns[7];
    ns[4].right = NULL;
    ns[7].parent = &ns[4];

    ns[5].left = &ns[8];
    ns[5].right = NULL;
    ns[8].parent = &ns[5];

    return ns;
}

template < typename T >
void printInOrder(BTreeNode<T>* node)
{
    if( node != NULL )
    {
        printInOrder(node->left);

        cout << node->value <<" ";

        printInOrder(node->right);
    }
}

template < typename T >
void printDualList(BTreeNode<T>* node)
{
    BTreeNode<T>* g = node;

    cout << "head -> tail: " << endl;

    while( node != NULL )
    {
        cout << node->value << " ";

        g = node;

        node = node->right;
    }

    cout << endl;

    cout << "tail -> head: " << endl;

    while( g != NULL )
    {
        cout << g->value << " ";

        g = g->left;
    }

    cout << endl;
}

template <typename T>
void inOrderThread2(BTreeNode<T> *node, BTreeNode<T> *& head, BTreeNode<T> *&tail)
{
    if (node != nullptr)
    {
        BTreeNode<T> *h = nullptr;
        BTreeNode<T> *t = nullptr;

        inOrderThread2(node->left, h, t);

        node->left = t;

        if (t != nullptr)
        {
            t->right = node;
        }

        head = (h != nullptr) ? h : node;

        h = nullptr;
        t = nullptr;

        inOrderThread2(node->right, h, t);

        node->right = h;

        if (h != nullptr)
        {
            h->left = node;
        }

        tail = (t != nullptr) ? t : node;
    }
}

template <typename T>
BTreeNode<T> *inOrderThread2(BTreeNode<T> *node)
{
    BTreeNode<T> *head = nullptr;
    BTreeNode<T> *tail = nullptr;

    inOrderThread2(node, head, tail);

    return head;
}

int main()
{
    BTreeNode<int>* ns = createTree<int>();

    printInOrder(ns);

    cout << endl;

    ns = inOrderThread2(ns);

    printDualList(ns);

    return 0;
}

    输出：

    3 6 1 0 7 4 2 8 5
head -> tail:
3 6 1 0 7 4 2 8 5
tail -> head:
5 8 2 4 7 0 1 6 3

    以上内容整理于狄泰软件学院系列课程，请大家保护原创！

    c++
    TianSong
    734 声望138 粉丝

    阿里山神木的种子在3000年前已经埋下，今天不过是看到当年注定的结果，为了未来的自己，今天就埋下一颗好种子吧

    « 上一篇
    【数据结构】69_二叉树的线索化实现
    下一篇 »
    【数据结构】71_图的定义与操作

    引用和评论

    推荐阅读
    头像
    《Linux系统编程训练营》1-2_多核调度预备知识

    TianSong赞 1阅读 96

    头像
    Linux C++ 开发6 - GDB调试入门指南

    陌尘赞 1阅读 616

    头像
    调用SCIP C++接口，viusal studio项目工程搭建（For新手）

    楚知行阅读 1.5k

    头像
    掌握Linux后端开发岗位技能，一本就够！！！

    后端开发工程实践阅读 1.3k评论 2

    头像
    VSCode系列 - 如何用VSCode搭建C++高效开发环境(2)

    陌尘阅读 1k

    头像
    CPP 小技巧

    轻口味赞 1阅读 153

    头像
    Linux C++ 开发4 - 入门makefile一篇文章就够了

    陌尘阅读 468

    0 条评论
    得票最新