leetcode-1006 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.

题意：根据中序遍历和后序遍历，构建二叉树

思路很清晰，做法很简单，就不讲了。

一开始我写了一个递归的解法，本地测试数据都OK，无奈提交的时候内存超出限制，下面先给出超出内存的代码：

 /**
  * 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) {
         if(inorder.size()==)
         return nullptr;
         if(inorder.size()==)
         return new TreeNode(inorder[]);
         int fath=postorder[postorder.size()-];
         TreeNode* root=new TreeNode(fath);
         int flag=-;
         for(int i=;i<inorder.size();i++)
         {
             if(inorder[i]==fath)
             {
                 flag=i;
                 break;
             }
         }
         vector<int> left;
         for(int i=;i<flag;i++)
         left.push_back(inorder[i]);
         vector<int> right;
         for(int i=flag+;i<inorder.size();i++)
         right.push_back(inorder[i]);
         int flag1=-;
         for(int i=;i<postorder.size();i++)
         {
             if(postorder[i]==inorder[flag])
             {
                 flag1=i;
                 break;
             }
         }

         vector<int> left1;
         for(int i=;i<flag1;i++)
         left1.push_back(postorder[i]);
         vector<int> right1;
         for(int i=flag1;i<postorder.size()-;i++)
         right1.push_back(postorder[i]);

         root->left=buildTree(left,left1);
         root->right=buildTree(right,right1);
         return root;
     }
 };

有没有看出问题，没错，就是第28、31、44、47行的代码，每次递归都会产生新的vector数组，所以最后导致内存超出限制。所以改进了一下，新定义一个方法helper来递归，helper里面的实现不再申请新的vector空间，直接在参数inorder和postorder中进行操作，从而避免内存超出限制。下面是accepted的代码：

 /**
  * 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 helper(inorder,,inorder.size()-,postorder,,postorder.size()-);
     }

     TreeNode* helper(vector<int>& inorder,int begin1,int end1,vector<int>& postorder,int begin2,int end2)
     {
         if(begin1>end1)
             return nullptr;
         if(begin1==end1)
             return new TreeNode(inorder[begin1]);

         TreeNode* root=new TreeNode(postorder[end2]);
         int i=begin1;
         for(;i<=end1;i++)
         {
             if(inorder[i]==postorder[end2])
             break;
         }
         int leftlen=i-begin1;

         root->left=helper(inorder,begin1,begin1+leftlen-,postorder,begin2,begin2+leftlen-);
         root->right=helper(inorder,begin1+leftlen+,end1,postorder,begin2+leftlen,end2-);
         return root;
     }
 };