【数据结构】70_二叉树的经典面试题分析

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

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

编程实验：单度结点的删除

#include 
#include "BTreeNode.h"

using namespace std;
using namespace DTLib;

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

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

        printInOrder(node->right);
    }
}

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

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

            if (parent != nullptr)
            {
                BTreeNode *& 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* ns = createTree();

    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

编程实验：单度结点删除

#include 
#include "BTreeNode.h"

using namespace std;
using namespace DTLib;

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

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

        printInOrder(node->right);
    }
}

template < typename T >
void printDualList(BTreeNode* node)
{
    BTreeNode* 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 
void delOdd2(BTreeNode *&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* ns = createTree();

    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

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

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

编程实验：中序线索化

#include 
#include "BTreeNode.h"

using namespace std;
using namespace DTLib;

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

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

        printInOrder(node->right);
    }
}

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

        node->left = pre;

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

        pre = node;

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

template 
BTreeNode *inOrderThread(BTreeNode *node)
{
    BTreeNode * pre = nullptr;

    inOrderThread(node, pre);

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

    return node;
}

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

    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

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

编程实验：中序线索化

#include 
#include "BTreeNode.h"

using namespace std;
using namespace DTLib;

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

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

        printInOrder(node->right);
    }
}

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

    inOrderThread2(node, head, tail);

    return head;
}

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

    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

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