重建二叉树

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题目描述

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

#include<iostream>#include"BinaryTree.h"using namespace std;BinaryTreeNode* ConstructTreeCore(int* startPreOrder, int* endPreOrder, int* startInOrder, int* endInOrder ){int rootValue = *startPreOrder;BinaryTreeNode* root = new BinaryTreeNode();root->m_nValue = rootValue;root->m_pLeft = root->m_pRight = NULL;if(startPreOrder == endPreOrder ){if(startInOrder == endInOrder && *startPreOrder == *startInOrder)return root;else{throw exception("Invaliad input");}}int *rootInOrder = startInOrder;while (rootInOrder < endInOrder && *rootInOrder != rootValue)rootInOrder++;if(rootInOrder == endInOrder && *rootInOrder != rootValue)throw exception("Invalid input");int leftLength = rootInOrder - startInOrder;int* leftPreNode = startPreOrder + leftLength;if(leftLength > 0)root ->m_pLeft = ConstructTreeCore(startPreOrder+1, leftPreNode,  startInOrder, rootInOrder-1 );if(endInOrder - startInOrder > leftLength)root ->m_pRight = ConstructTreeCore(leftPreNode+1, endPreOrder,  rootInOrder+1, endInOrder );return root;}BinaryTreeNode* ConstructTree(int* preOrder, int* inOrder,int length){if(preOrder == NULL || inOrder == NULL || length <=0)return NULL;int* startPreOrder =  preOrder;int* endPreOrder = preOrder + length-1;int* startInOrder = inOrder;int* endInOrder = inOrder + length-1;return ConstructTreeCore(startPreOrder, endPreOrder, startInOrder, endInOrder );}

// ====================测试代码====================void Test(char* testName, int* preorder, int* inorder, int length){    if(testName != NULL)        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 = ConstructTree(preorder, inorder, length);        PrintTree(root);        DestroyTree(root);    }    catch(exception& exception)    {        printf("Invalid Input.\n");    }}// 普通二叉树//              1//           /     \//          2       3  //         /       / \//        4       5   6//         \         ///          7       8void 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//     ///    5void 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//                   \//                    5void 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   7void 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", NULL, NULL, 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);}int main(int argc, char* argv[]){    Test1();    Test2();    Test3();    Test4();    Test5();    Test6();    Test7();    return 0;}

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