105. 从前序与中序遍历序列构造二叉树

根据一棵树的前序遍历与中序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如，给出

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]

返回如下的二叉树：

    3
   / \
  9  20
    /  \
   15   7

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector& preorder, vector& inorder) {
        if(!preorder.size()) return NULL;
        int middle = preorder[0];
        vector preleft;
        vector preright;
        vector inleft;
        vector inright;
        int i, j;
        i = 0;
        j = 1;
        for(;i < inorder.size();i++){
            if(inorder[i] != middle) inleft.push_back(inorder[i]);
            else break;
        }
        i++;
        for(;i < inorder.size();i++){
            inright.push_back(inorder[i]);
        }
        for(;j <= inleft.size();j++){
            preleft.push_back(preorder[j]);
        }
        for(;j < preorder.size();j++){
            preright.push_back(preorder[j]);
        }
        TreeNode* T = new TreeNode(middle);
        T->left = buildTree(preleft, inleft);
        T->right = buildTree(preright, inright);
        return T;
    }
};