解题思路：

思路和Leetcode 105题相同。区别在于，在这一题中，后序遍历的最后一个值为根节点。

然后仍然是找到根节点后，划分左右子树，递归构建。

/**
 * 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) {
        return creat(inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1);
    }
    TreeNode* creat(vector<int>& inorder, int il, int ir, vector<int>& postorder, int pl, int pr){
        if(il > ir || pl > pr) return NULL;
        TreeNode* root = new TreeNode(postorder[pr]);
        for(int i = il; i <= ir; i++){
            if(postorder[pr] == inorder[i]){
                root->left = creat(inorder, il, i-1, postorder, pl, pl+i-il-1);
                root->right = creat(inorder, i+1, ir, postorder, pl+i-il, pr-1);
                break;
            }
        }
        return root;
    }
};