73前序遍历和中序遍历树构造二叉树

根据前序遍历和中序遍历树构造二叉树.

给出中序遍历：[1,2,3]和前序遍历：[2,1,3]. 返回如下的树:

  2 / \1   3

解题思路：1.先序遍历的第一个节点为根节点。2.在中序遍历中定位到根节点的位置，左半部分即为左子树，右半部分即为右子树。3.利用递归思想重构二叉树。

已AC的两种代码：
/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } */ class Solution {    /**     *@param preorder : A list of integers that preorder traversal of a tree     *@param inorder : A list of integers that inorder traversal of a tree     *@return : Root of a tree     */public:    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {       if(preorder.size()==0||inorder.size()==0)        {        return NULL;    }   TreeNode*sroot=new TreeNode(preorder[0]);   sroot->left=NULL;        sroot->right=NULL;        int pos=-1;        vector<int>preleft;        vector<int>preright;        vector<int>inleft;        vector<int>inright;        int sl=0;        int sr=0;   for(int i=0;i<inorder.size();i++)            {            if(inorder[i]==preorder[0])                {                pos=i;                break;            }        }        for(int i=0;i<inorder.size();i++)            {            if(i<pos)                {                inleft.push_back(inorder[i]);                sl++;            }            if(i>pos)                {                inright.push_back(inorder[i]);                sr++;            }        }        for(int i=1;i<preorder.size();i++)            {            if(sl)                {                preleft.push_back(preorder[i]);                --sl;                continue;            }            if(sr)                {                preright.push_back(preorder[i]);                --sr;            }        }        sroot->left=buildTree(preleft,inleft);        sroot->right=buildTree(preright,inright);        return sroot; // write your code here    }};
/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } */class Solution {    /**     *@param preorder : A list of integers that preorder traversal of a tree     *@param inorder : A list of integers that inorder traversal of a tree     *@return : Root of a tree     */public:    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {       if(preorder.size()==0||inorder.size()==0)        {        return NULL;    }   TreeNode*sroot=new TreeNode(preorder[0]);   sroot->left=NULL;        sroot->right=NULL;        vector<int>preleft;        vector<int>preright;        vector<int>inleft;        vector<int>inright;        int pos=find(inorder.begin(),inorder.end(),preorder[0])-inorder.begin();        inleft.assign(inorder.begin(),inorder.begin()+pos);        inright.assign(inorder.begin()+pos+1,inorder.end());        preleft.assign(preorder.begin()+1,preorder.begin()+pos+1);        preright.assign(preorder.begin()+pos+1,preorder.end());        sroot->left=buildTree(preleft,inleft);        sroot->right=buildTree(preright,inright);        return sroot; // write your code here    }};