剑指offer---面试题6重建二叉树

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。重建出二叉树，并输出根节点。二叉树的定义如下：

struct BinaryTreeNode
{
int m_nValue;
struct BinaryTreeNode * m_pLeft;
struct BinaryTreeNode * m_pRigth;
};

如上，前序遍历（1,2,4,7,3,5,6,8,），中序（4,7,2,1,5,3，8,6,）。后序遍历（7,4,2,5，8，6,3,1）

分析：

在二叉树的前序遍历中，第一个数字总是根节点，单在中序中，左子树的节点位于根节点的左侧，右子树在右侧。

可以用递归方法，完成分别构建左子树和右子树。

//已知前序和中序遍历，构造二叉树。///#include "stdafx.h"#include <stdio.h>#include <exception>struct BinaryTreeNode{int m_nValue;struct BinaryTreeNode * m_pLeft;struct BinaryTreeNode * m_pRigth;};BinaryTreeNode * ConstructCore(int *startPreorder,int *endPreorder,int *startInorder,int *endInorder){//the first node in preorder is the root nodeint rootValue=startPreorder[0];BinaryTreeNode *root = new BinaryTreeNode();root->m_nValue=rootValue;root->m_pLeft=root->m_pRigth=NULL;//end conditionif(startPreorder==endPreorder){if(*startPreorder==*endPreorder && startInorder==endInorder)return  root;elsethrow std::exception("invalid input");}//在中序遍历找到根节点的值int *rootInorder = startInorder;while(*rootInorder!=rootValue && rootInorder<=endInorder)rootInorder++;if(*rootInorder!=rootValue && rootInorder==endInorder)throw std::exception("invalid input!");int leftLength=rootInorder-startInorder;int *leftPreorderEnd = startPreorder+leftLength;if(leftLength>0){root->m_pLeft = ConstructCore(startPreorder+1,leftPreorderEnd,startInorder,rootInorder-1);}if(leftLength<endPreorder-startPreorder)root->m_pRigth=ConstructCore(leftPreorderEnd+1,endPreorder,rootInorder+1,endInorder);return root;}BinaryTreeNode * Construct(int *preorder,int *inorder,int length){if(preorder==NULL || inorder==NULL || length<=0)return 0;return ConstructCore(preorder,preorder+length-1,inorder,inorder+length-1);}/////test codevoid PrintTreeNode(BinaryTreeNode *pNode){if(pNode!=NULL){printf("this node value is : %d\n",pNode->m_nValue);if(pNode->m_pLeft != NULL)            printf("value of its left child is: %d.\n", pNode->m_pLeft->m_nValue);        else            printf("left child is null.\n");        if(pNode->m_pRigth != NULL)            printf("value of its right child is: %d.\n", pNode->m_pRigth->m_nValue);        else            printf("right child is null.\n");}elseprintf("this node is NULL.\n");printf("\n");}void PrintTree(BinaryTreeNode *pRoot){PrintTreeNode(pRoot);if(pRoot!=NULL){if(pRoot->m_pLeft!=NULL)PrintTree(pRoot->m_pLeft);if(pRoot->m_pRigth!=NULL)PrintTree(pRoot->m_pRigth);}}void DestoryTree(BinaryTreeNode *pRoot){if(pRoot!=NULL){BinaryTreeNode * pLeft=pRoot->m_pLeft;BinaryTreeNode *pRight = pRoot->m_pRigth;delete pRoot;pRoot=NULL;DestoryTree(pLeft);DestoryTree(pRight);}}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(i = 0; i < length; ++ i)        printf("%d ", inorder[i]);    printf("\n");    try    {        BinaryTreeNode* root = Construct(preorder, inorder, length);        PrintTree(root);        DestoryTree(root);    }    catch(std::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();printf("Hello World!\n");return 0;}