4.重建二叉树（做第二遍时感觉仍有有难度，第三次做还是要看一下思路才行）

重建二叉树

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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}，则重建二叉树并返回。

我没有按照书上的做法来解，而是采用long型指针来存储地址，虽然很绕，但是可以熟悉指针的使用。牛客网使用的是64位编译器，故不能用int数组存地址，而要使用long型数组。

<pre name="code" class="cpp">// 4.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <vector>using namespace::std;struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:struct TreeNode* reConstructBinaryTree(vector<int> pre, vector<int> in) {if (pre.size() <= 0 || in.size() <= 0) return NULL;if (pre.size() != in.size()) return NULL;int length = pre.size(); // 获得数组的长度long* pPre = new long[length];int i = 0;for (vector<int>::iterator iter = pre.begin(); iter != pre.end(); iter++){pPre[i] = (long)&(*iter);// *iter取迭代器中的内容，&(*iter)取该内容的地址，(long)&(*iter)把地址转化为long型i++;}long* pIn = new long[length]; // 动态分配数组内存i = 0;for (vector<int>::iterator iter = in.begin(); iter != in.end(); iter++){pIn[i] = (long)&(*iter);i++;}TreeNode* root = ConstructCore(pPre, pPre + length - 1, pIn, pIn + length - 1);// 释放堆上的内存，防止内存泄漏，并清空指针delete[] pPre;pPre = NULL;delete[] pIn;pIn = NULL;return root;}private:TreeNode* ConstructCore(long* pPreStart, long* pPreEnd, long* pInStart, long* pInEnd){// 构建根节点long rootValue = *(long*)pPreStart[0];TreeNode* root = new TreeNode(rootValue);root->left = root->right = NULL;// 判断是否为递归的退出条件：前序是否到头、中序是否到头、前序与中序是否重合if (pPreStart == pPreEnd || pInStart == pInEnd || pPreStart == pInStart){return root;}// 在中序数组中找到根节点的位置long* rootPosi = pInStart;while (rootPosi <= pInEnd && *(long*)rootPosi[0] != rootValue){rootPosi++;}// 检验是否在中序数组中找到根节点if (rootPosi == pInEnd && *(long*)rootPosi[0] != rootValue){ // *(long*)rootPosi与*(long*)rootPosi[0]意义完全不同，*(long*)rootPosi是把数组的第一个元素到最后一个元素看成整体解除指针引用//throw new exception("not found!");return NULL;}// 中序数组中根节点据数组头部的距离long leftLength = rootPosi - pInStart;// 构建左子树的四个实参long* pNewPreStartLeft = pPreStart + 1;long* pNewPreEndLeft = pPreStart + leftLength;long* pNewInStartLeft = pInStart;long* pNewInEndLeft = pInStart + leftLength - 1;// 构建右子树的四个实参long* pNewPreStartRight = pPreStart + leftLength + 1;long* pNewPreEndRight = pPreEnd;long* pNewInStartRight = rootPosi + 1;long* pNewInEndRigth = pInEnd;if (leftLength > 0){root->left = ConstructCore(pNewPreStartLeft, pNewPreEndLeft, pNewInStartLeft, pNewInEndLeft);}if (leftLength < pPreEnd - pPreStart){root->right = ConstructCore(pNewPreStartRight, pNewPreEndRight, pNewInStartRight, pNewInEndRigth);}return root;}};int _tmain(int argc, _TCHAR* argv[]){int arrPre[] = { 1, 2, 4, 7, 3, 5, 6, 8 };int arrIn[] = { 4, 7, 2, 1, 5, 3, 8, 6 };vector<int> pre(arrPre, arrPre + 8);vector<int> in(arrIn, arrIn + 8);Solution s;TreeNode* result = s.reConstructBinaryTree(pre, in);system("pause");return 0;}

第二次做：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:struct TreeNode* reConstructBinaryTree(vector<int> pre, vector<int> in) {        if ( pre.empty() || in.empty() || pre.size() != in.size() ) return NULL ;                int length = pre.size() ;                TreeNode* root = new TreeNode( pre[0] ) ;                int rootVal = pre[0] ;        int mid = 0;        for ( ; in[mid] != rootVal; ++ mid ) {            if ( mid > length) return NULL ;        }                vector<int> pre_left ;        vector<int> in_left ;        for ( int i = 0; i < mid; ++ i ) {            pre_left.push_back( pre[i + 1] ) ;            in_left.push_back( in[i] ) ;        }                vector<int> pre_right ;        vector<int> in_right ;        for ( int i = mid + 1; i < length; ++ i ) {            pre_right.push_back( pre[i] ) ;            in_right.push_back( in[i] ) ;        }                root -> left = reConstructBinaryTree( pre_left, in_left ) ;        root -> right = reConstructBinaryTree( pre_right, in_right ) ;                return root ;}};

第三次做：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    struct TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> in) {if ( pre.empty() || in.empty() || pre.size() != in.size() ) return NULL ;                TreeNode* root = new TreeNode( pre[0] ) ;                int rootVal = pre[ 0 ] ;        int mid = 0 ;        int length = pre.size() ;                for ( ; in[mid] != rootVal; ++ mid ) {           // if ( mid > length ) return NULL ;        }                vector<int> pre_left ;        vector<int> in_left ;        for ( int i = 0; i < mid; ++ i ) {            pre_left.push_back( pre[i + 1] ) ;            in_left.push_back( in[i] ) ;        }                vector<int> pre_right ;        vector<int> in_right ;        for ( int i = mid + 1; i < length; ++ i ) {            pre_right.push_back( pre[i] ) ;            in_right.push_back( in[i] ) ;        }                root->left = reConstructBinaryTree( pre_left, in_left ) ;        root->right = reConstructBinaryTree( pre_right, in_right ) ;                return root ;    }};