LeetCode_Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

　　

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode * constructSubtree(vector<int> &inorder, int inStart, int inEnd, vector<int> &postorder, int postStart, int postEnd)
    {
        int temp = postorder[postEnd] ;
        TreeNode * root = new TreeNode(temp) ;
    
        if(inStart == inEnd ) return root;
    
        int i ;
        for( i = inStart ; i<= inEnd ; i++)
        if( inorder[i] == temp) break;
      
        root->left = inStart <= i-1 ? constructSubtree(inorder, inStart, i-1, postorder,postStart, postStart + i - inStart - 1  ) : NULL;
    
        root->right = i+1 <= inEnd ? constructSubtree(inorder, i+1, inEnd, postorder, postStart + i - inStart, postEnd -1 ) : NULL ;
 
        return root ;
 
    }
 
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(inorder.size() == 0) return NULL;
        int temp = postorder[postorder.size()-1] ;
        TreeNode * root = new TreeNode(temp) ;
        
        int i;
        for(i = 0; i< inorder.size(); i++)
         if(inorder[i] == temp) break ;
         
        root->left = 0 <= i-1 ? constructSubtree(inorder, 0, i-1, postorder, 0, i-1) : NULL ;
        root->right =  i+1 <= inorder.size()- 1 ? constructSubtree(inorder, i+1, inorder.size()-1, postorder, i, postorder.size() -2) : NULL;
        
        return root ;
    }
};