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}，则重建出二叉树并输出它的头结点。二叉树结点的定义如下：

题目分析

剑指Offer（纪念版）P55

代码实现

分治与递归

BinaryTreeNode* Construct(int* preorder, int* inorder, int length){    if(preorder == NULL || inorder == NULL || length <= 0)        return NULL;    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 = NULL;    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;}