重建二叉树

题目：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{ 1,2, 4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}则重建出图2.6所示的二叉树并输出它的头结点。

#pragma once
#include<queue>
struct BTNode
{
int value;
BTNode* left;
BTNode* right;
};

BTNode* ConstructCore(int * startPrevorder, int* endPreorder ,
int* startInorder , int* endInorder)
{
int rootValue = startPrevorder [0];
BTNode *root = new BTNode();
root->value = rootValue;
root->left = root->right = NULL;
if (startPrevorder == endPreorder)
{
if (startInorder == endInorder
&&* startPrevorder == *startInorder )
return root;
else
throw std::exception ("Invalid input.");
}
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 = startPrevorder + leftLength;
if (leftLength > 0)
{
root->left = ConstructCore( startPrevorder + 1,
leftPreorderEnd, startInorder, rootInorder - 1);

}
if (leftLength < endPreorder - startPrevorder)
{
root->right = ConstructCore(leftPreorderEnd + 1,
endPreorder, rootInorder + 1, endInorder );
}
return root;
}

BTNode* Construct(int * prevorder, int* inorder , int length)
{
if (prevorder == NULL || inorder == NULL || length <= 0)
return NULL ;
return ConstructCore(prevorder , prevorder + length - 1,
inorder, inorder + length - 1);
}

void Test()
{
int pre[8] = { 1, 2, 4, 7, 3, 5, 6, 8 };
int in[8] = { 4, 7, 2, 1, 5, 3, 8, 6 };

BTNode* root = Construct(pre, in, 8);

//PrintTree(root);
}