/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* get_tree(vector<int>& inorder, vector<int>& postorder)
    {
      //如果中序数组或者后序数组是空，就返回空节点
        if(inorder.size()==0||postorder.size()==0) return nullptr;
    //后续数组的最后一个是根节点。
        TreeNode* root = new TreeNode(postorder[postorder.size()-1]);
    //根据根节点的值，去找到对应的中序根节点
        int root_num ;
        for(int i= 0 ;i<inorder.size();i++)
        {
            if(inorder[i] == root->val)
            {
                root_num = i;
                break;
            }
        }
    //通过中序数组中根节点的位置，根节点前面是左子树，后面是右子树
        vector<int> inorder_left(inorder.begin() , inorder.begin()+root_num);
        vector<int> inorder_right(inorder.begin()+root_num+1 , inorder.end());
    //通过中序分割出左子树和右子树的长度，去分割后续数组的左子树和右子树
        vector<int> postorder_left(postorder.begin() , postorder.begin() + inorder_left.size());
        vector<int> postorder_right(postorder.begin()+ inorder_left.size() , postorder.end()-1);
    //然后递归迭代，左子树的中序和后续，右子树的中序和后续
        root->left =  get_tree(inorder_left , postorder_left);
        root->right = get_tree(inorder_right , postorder_right );
    //返归根节点
        return root;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
    //如果输入两个数组是空，则返回空节点
        if(inorder.size()==0||postorder.size()==0) return nullptr;
        return get_tree(inorder, postorder);
    }
};

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* track_back(vector<int>& inorder, vector<int>& postorder)
    {
        if(inorder.size()==0 || postorder.size() == 0) return nullptr;
        int rootValue = postorder[postorder.size()-1];
        TreeNode *root = new TreeNode(rootValue);
        if(postorder.size()==1 ) return root;
        int delimiter ;
        for(delimiter=0 ; delimiter < inorder.size();delimiter++)
            if(inorder[delimiter] == rootValue) break;
        vector<int> leftInorder(inorder.begin() , inorder.begin() + delimiter);
        vector<int> rightInorder(inorder.begin() + delimiter + 1 , inorder.end());
         postorder.erase(postorder.end()-1);
        // postorder.resize(postorder.size()-1);
        vector<int> leftPostorder(postorder.begin() , postorder.begin() + leftInorder.size());
        vector<int> rightPostorder(postorder.begin()+leftInorder.size() , postorder.end());
        root->left = track_back(leftInorder,leftPostorder);
        root->right = track_back(rightInorder,rightPostorder);
        return root;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if(inorder.size()==0 || postorder.size() == 0) return nullptr;
        TreeNode *root =  track_back( inorder ,postorder);
        return root;
    }
};