重建二叉树

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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{ 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);}

#include<iostream>using namespace std;#include<vector>#include <algorithm> struct TreeNode{int _value;TreeNode* _left;TreeNode* _right;TreeNode(const int x):_value(x), _left(NULL), _right(NULL){}};TreeNode* reConstructBinaryTree(vector<int> pre, vector<int> in) {TreeNode* tree = new TreeNode(pre[0]);//tree->val = pre.front();vector<int>::iterator lt = find(in.begin(), in.end(), tree->_value);//lt指向1//中序中it+1指向vector<int>::iterator rt = find_first_of(pre.begin(), pre.end(), lt + 1, in.end());//rt指向3vector<int>l1(pre.begin() + 1, rt);//2 4 7 迭代器左闭右开，所以没有3vector<int>r1(in.begin(), lt);//4  7 2 vector<int>l2(rt, pre.end()); //3 5 6 8vector<int>r2(lt + 1, in.end());//5 3 8 6if (l1.empty()) tree->_left = NULL;else//继续进行递归tree->_left = reConstructBinaryTree(l1, r1);//l1和r1全为左树的if (l2.empty())tree->_right = NULL;else //继续进行递归tree->_right = reConstructBinaryTree(l2, r2);//l2和r2全为右树情况return tree;}int main(){int prev[] = { 1, 2, 4, 7, 3, 5, 6, 8 };vector<int> pre(prev, prev + sizeof(prev) / sizeof(int));int ins[] = { 4, 7, 2, 1, 5, 3, 8, 6 };vector<int> in(ins, ins + sizeof(ins) / sizeof(int));TreeNode* root = reConstructBinaryTree( pre, in);system("pause");return 0;}

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