To prove safety notes offer face questions 7 ---- rebuild binary tree

Title: and enter the result in a preorder traversal of a binary tree in preorder traversal of the binary tree rebuild. Suppose Results preorder traversal order and input of duplicate numbers are free. For example, the input preorder traversal sequence {1, 2, 4, 7, 3, 5, 6, 8}, and preorder sequence {4, 7, 2, 1, 53, 8, 6}, the reconstruction below and outputting the binary tree shown in its head node. Node binary tree defined as follows:

struct BinaryTreeNode{
    int m_nValue;
    BinaryTreeNode* m_pLeft;
    BinaryTreeNode* m_pRight;
};
//              1
//           /     \
//          2       3  
//         /       / \
//        4       5   6
//         \         /
//          7       8

Test Case:

• Ordinary binary tree (complete binary tree, not a complete binary tree).
• Special binary tree (all nodes have no right child of a binary tree; all nodes are not left child node of a binary tree; only one node of a binary tree).
• Special test input (nullptr a binary tree root pointer; input preamble sequence preorder traversal sequence does not match).

Test code:

void Test(char* testName, int* preorder, int* inorder, int length)
{
    if(testName != nullptr)
        printf("%s begins:\n", testName);
    printf("The preorder sequence is: ");
    for(int i = 0; i < length; ++ i)
        printf("%d ", preorder[i]);
    printf("\n");
    printf("The inorder sequence is: ");
    for(int i = 0; i < length; ++ i)
        printf("%d ", inorder[i]);
    printf("\n");
    try
    {
        BinaryTreeNode* root = Construct(preorder, inorder, length);
        PrintTree(root);
        DestroyTree(root);
    }
    catch(std::exception& exception)
    {
        printf("Invalid Input.\n");
    }
}

// 普通二叉树
//              1
//           /     \
//          2       3  
//         /       / \
//        4       5   6
//         \         /
//          7       8
void Test1()
{
    const int length = 8;
    int preorder[length] = {1, 2, 4, 7, 3, 5, 6, 8};
    int inorder[length] = {4, 7, 2, 1, 5, 3, 8, 6};
    Test("Test1", preorder, inorder, length);
}

// 所有结点都没有右子结点
//            1
//           / 
//          2   
//         / 
//        3 
//       /
//      4
//     /
//    5
void Test2()
{
    const int length = 5;
    int preorder[length] = {1, 2, 3, 4, 5};
    int inorder[length] = {5, 4, 3, 2, 1};
    Test("Test2", preorder, inorder, length);
}

// 所有结点都没有左子结点
//            1
//             \ 
//              2   
//               \ 
//                3 
//                 \
//                  4
//                   \
//                    5
void Test3()
{
    const int length = 5;
    int preorder[length] = {1, 2, 3, 4, 5};
    int inorder[length] = {1, 2, 3, 4, 5};
    Test("Test3", preorder, inorder, length);
}

// 树中只有一个结点
void Test4()
{
    const int length = 1;
    int preorder[length] = {1};
    int inorder[length] = {1};
    Test("Test4", preorder, inorder, length);
}

// 完全二叉树
//              1
//           /     \
//          2       3  
//         / \     / \
//        4   5   6   7
void Test5()
{
    const int length = 7;
    int preorder[length] = {1, 2, 4, 5, 3, 6, 7};
    int inorder[length] = {4, 2, 5, 1, 6, 3, 7};
    Test("Test5", preorder, inorder, length);
}

// 输入空指针
void Test6()
{
    Test("Test6", nullptr, nullptr, 0);
}

// 输入的两个序列不匹配
void Test7()
{
    const int length = 7;
    int preorder[length] = {1, 2, 4, 5, 3, 6, 7};
    int inorder[length] = {4, 2, 8, 1, 6, 3, 7};
    Test("Test7: for unmatched input", preorder, inorder, length);
}

This question test sites:

• Comprehension test candidates of the binary tree in preorder traversal and traversing. Only different binary tree traversal algorithm to have a deep understanding of possible candidates left in the divided traversal sequence, the sequence corresponding to the right child.
• The ability to examine the complex issues of the candidates analysis. We build big problem is broken down into binary construct two small problems left and right sub-tree. We found a small problem and a big problem in essence is the same, and therefore can be solved with a recursive manner.

Implementation code:

/*********************************BinaryTree.h************************************/
struct BinaryTreeNode 
{
    int                    m_nValue; 
    BinaryTreeNode*        m_pLeft;  
    BinaryTreeNode*        m_pRight; 
};

BinaryTreeNode* CreateBinaryTreeNode(int value);
void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, BinaryTreeNode* pRight);
void PrintTreeNode(const BinaryTreeNode* pNode);
void PrintTree(const BinaryTreeNode* pRoot);
void DestroyTree(BinaryTreeNode* pRoot);

/*********************************BinaryTree.cpp************************************/
#include <cstdio>
#include "BinaryTree.h"

BinaryTreeNode* CreateBinaryTreeNode(int value)
{
    BinaryTreeNode* pNode = new BinaryTreeNode();
    pNode->m_nValue = value;
    pNode->m_pLeft = nullptr;
    pNode->m_pRight = nullptr;
    return pNode;
}

void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, BinaryTreeNode* pRight)
{
    if(pParent != nullptr)
    {
        pParent->m_pLeft = pLeft;
        pParent->m_pRight = pRight;
    }
}

void PrintTreeNode(const BinaryTreeNode* pNode)
{
    if(pNode != nullptr)
    {
        printf("value of this node is: %d\n", pNode->m_nValue);

        if(pNode->m_pLeft != nullptr)
            printf("value of its left child is: %d.\n", pNode->m_pLeft->m_nValue);
        else
            printf("left child is nullptr.\n");

        if(pNode->m_pRight != nullptr)
            printf("value of its right child is: %d.\n", pNode->m_pRight->m_nValue);
        else
            printf("right child is nullptr.\n");
    }
    else
    {
        printf("this node is nullptr.\n");
    }
    printf("\n");
}

void PrintTree(const BinaryTreeNode* pRoot)
{
    PrintTreeNode(pRoot);

    if(pRoot != nullptr)
    {
        if(pRoot->m_pLeft != nullptr)
            PrintTree(pRoot->m_pLeft);
        if(pRoot->m_pRight != nullptr)
            PrintTree(pRoot->m_pRight);
    }
}

void DestroyTree(BinaryTreeNode* pRoot)
{
    if(pRoot != nullptr)
    {
        BinaryTreeNode* pLeft = pRoot->m_pLeft;
        BinaryTreeNode* pRight = pRoot->m_pRight;
        delete pRoot;
        pRoot = nullptr;
        DestroyTree(pLeft);
        DestroyTree(pRight);
    }
}

/*********************************ConstructBinaryTree.cpp************************************/
#include "..\Utilities\BinaryTree.h"
#include <exception>
#include <cstdio>

BinaryTreeNode* ConstructCore(int* startPreorder, int* endPreorder, int* startInorder, int* endInorder);

BinaryTreeNode* Construct(int* preorder, int* inorder, int length)
{
    if(preorder == nullptr || inorder == nullptr || length <= 0)
        return nullptr;
    return ConstructCore(preorder, preorder + length - 1,
        inorder, inorder + length - 1);
}

BinaryTreeNode* ConstructCore
(
    int* startPreorder, int* endPreorder, 
    int* startInorder, int* endInorder
)
{
    // 前序遍历序列的第一个数字是根结点的值
    int rootValue = startPreorder[0];
    BinaryTreeNode* root = new BinaryTreeNode();
    root->m_nValue = rootValue;
    root->m_pLeft = root->m_pRight = nullptr;

    if(startPreorder == endPreorder)
    {
        if(startInorder == endInorder && *startPreorder == *startInorder)
            return root;
        else
            throw std::exception("Invalid input.");
    }

    // 在中序遍历中找到根结点的值
    int* rootInorder = startInorder;
    while(rootInorder <= endInorder && *rootInorder != rootValue)
        ++ rootInorder;
    if(rootInorder == endInorder && *rootInorder != rootValue)
        throw std::exception("Invalid input.");
    int leftLength = rootInorder - startInorder;
    int* leftPreorderEnd = startPreorder + leftLength;
    if(leftLength > 0)
    {
        // 构建左子树
        root->m_pLeft = ConstructCore(startPreorder + 1, leftPreorderEnd, 
            startInorder, rootInorder - 1);
    }
    if(leftLength < endPreorder - startPreorder)
    {
        // 构建右子树
        root->m_pRight = ConstructCore(leftPreorderEnd + 1, endPreorder,
            rootInorder + 1, endInorder);
    }
    return root;
}
int main(int argc, char* argv[])
{
    Test1();
    Test2();
    Test3();
    Test4();
    Test5();
    Test6();
    Test7();
    int a;
    scanf("%d", &a);
    return 0;
}