根据前序遍历和中序遍历还原二叉树

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/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */#include<assert.h>class Solution {public:    struct TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> in)    {        if (pre.empty() || in.empty()||pre.size()!=in.size())               return NULL;        int size = pre.size();        vector<int> left_pre,left_in,right_pre,right_in;        TreeNode* root = new TreeNode(pre[0]);        int i = 0;        for(;i<size;i++)        {         if(in[i] == pre[0])            break;        }        if(i == size)            assert(0);        for(int j =0; j < i;j++)        {           left_in.push_back(in[j]);           left_pre.push_back(pre[j+1]);        }        for(int j=i+1;j<size;++j)        {           right_in.push_back(in[j]);           right_pre.push_back(pre[j]);        }        if(!left_in.empty())        {           root->left =  reConstructBinaryTree(left_pre,left_in);        }        if(!right_in.empty())        {            root->right = reConstructBinaryTree(right_pre, right_in);        }        return root;    }};

template<typename T>struct BinaryTreeNode{    BinaryTreeNode(T data)    : _data(data)    , _pLeft(NULL)    , _pRight(NULL)    {}    T _data;    BinaryTreeNode<T>* _pLeft;    BinaryTreeNode<T>* _pRight;};

BinaryTree() :_proot(NULL){}    BinaryTreeNode<T>* reConstructBinaryTree(char* prev, char * in, int len)    {        if (prev == NULL || in == NULL || len <= 0)            return NULL;        _proot = _reConstructBinaryTree(prev, prev + len - 1, in, in + len - 1);        return _proot;    }BinaryTreeNode<T>* _reConstructBinaryTree(char* prev, \        char* prev_end, char * in, char* in_end)    {        BinaryTreeNode<T>* root = new BinaryTreeNode<T>(prev[0]);        if (prev == prev_end)        {            if (in == in_end&&*prev == *in)                return root;            else                assert(0);        }        char* pin = in;        while (pin <= in_end)//找到中序遍历的跟节点        {            if (*pin == *prev)                break;            else            {                pin++;            }        }        if (pin == in_end&& *pin != *prev)            assert(0);        int leftlength = pin - in;        char *leftprevend = prev + leftlength;        if (leftlength > 0)//构建左子树        {            root->_pLeft = _reConstructBinaryTree(prev + 1, leftprevend, in, pin - 1);        }        if (leftlength < prev_end - prev)//构建右子树        {            root->_pRight = _reConstructBinaryTree(leftprevend + 1, prev_end, pin + 1, in_end);        }        return root;    }

最后简单说一下思路，很简单。
1.前序的第一个是根
2.中序里找到根，前面的是左子树，后面的是右子树
3.递归的对左子树和右子树进行上述操作，
注意递归出口

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