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.

常规方法，以中序为主，在中序中查找根，划分成左右两部分递归，注意要记录左部分个数，后续遍历找根需要，这里没有考虑不能建成树的情况

/**
 * 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 *buildTree(vector<int> &inorder, vector<int> &postorder) {
        int m=inorder.size();
        int n=postorder.size();
        if(m==0||n==0||m!=n) return NULL;     //如果m，n不等，一定不对
        
        return build(inorder,0,n-1,postorder,0,n-1);
      
    }
    TreeNode * build(vector<int> & inorder,int inl,int inr,vector<int> &postorder,int pol,int por)
    {

        if(inl>inr)
        {
            return NULL;
        }
        int value=postorder[por];
        TreeNode * root=new TreeNode(value);
        int i;
        int count=0;                             //记录左部分个数                     
        for(i=inl;i<=inr;i++)
        {
            if(inorder[i]==value)
            {
                break;
            }
            count++;
        }
        root->left=build(inorder,inl,i-1,postorder,pol,pol+count-1);                   //注意count-1
        root->right=build(inorder,i+1,inr,postorder,pol+count,por-1);
        return root;
        
    }
};