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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。例如输入前序遍历序列为{1,2,4,7,3,5,6,8}和中序遍历的序列{4,7,2,1,5,3,8,6}，则重建出下图所示的二叉树并输出它的头结点。
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;elsethrow 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;}
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原文地址：http://www.cnblogs.com/zhuzhenfeng/p/4663539.html
