106. 从中序与后序遍历序列构造二叉树

根据一棵树的中序遍历与后序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如，给出

中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]

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<int>& inorder, vector<int>& postorder) {
        if(!postorder.size()) return NULL;
        int middle = postorder[postorder.size()-1];
        vector<int> postleft;
        vector<int> postright;
        vector<int> inleft;
        vector<int> inright;
        int i, j;
        i = j = 0;
        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++){
            postleft.push_back(postorder[j]);
        }
        for(;j < postorder.size()-1;j++){
            postright.push_back(postorder[j]);
        }
        TreeNode* T = new TreeNode(middle);
        T->left = buildTree(inleft, postleft);
        T->right = buildTree(inright, postright);
        return T;

    }
};