給定一棵樹的中序遍歷與後序遍歷，依據此構造二叉樹。
註意:
你可以假設樹中沒有重復的元素。
例如，給出
中序遍歷 = [9,3,15,20,7]
後序遍歷 = [9,15,7,20,3]
返回如下的二叉樹：
3
/ \
9 20
/ \
15 7
詳見：https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/

/**
 * 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) {
        int is=inorder.size();
        if(is==0||inorder.empty())
        {
            return nullptr;
        }
        int val=postorder[is-1];
        TreeNode* root=new TreeNode(val);
        vector<int> in_left,in_right,post_left,post_right;
        int p=0;
        for(;p<is;++p)
        {
            if(inorder[p]==val)
            {
                break;
            }
        }
        for(int i=0;i<is;++i)
        {
            if(i<p)
            {
                in_left.push_back(inorder[i]);
                post_left.push_back(postorder[i]);
            }
            else if(i>p)
            {
                in_right.push_back(inorder[i]);
                post_right.push_back(postorder[i-1]);
            }
        }
        root->left=buildTree(in_left,post_left);
        root->right=buildTree(in_right,post_right);
        return root;
    }
};

106 Construct Binary Tree from Inorder and Postorder Traversal 從中序與後序遍歷序列構造二叉樹