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()==0)
        return nullptr;
        if(inorder.size()==1)
        return new TreeNode(inorder[0]);
        int fath=postorder[postorder.size()-1];
        TreeNode* root=new TreeNode(fath);
        int flag=-1;
        for(int i=0;i<inorder.size();i++)
        {
            if(inorder[i]==fath)
            {
                flag=i;
                break;
            }
        }
        vector<int> left;
        for(int i=0;i<flag;i++)
        left.push_back(inorder[i]);
        vector<int> right;
        for(int i=flag+1;i<inorder.size();i++)
        right.push_back(inorder[i]);
        int flag1=-1;
        for(int i=0;i<postorder.size();i++)
        {
            if(postorder[i]==inorder[flag])
            {
                flag1=i;
                break;
            }
        }
        
        vector<int> left1;
        for(int i=0;i<flag1;i++)
        left1.push_back(postorder[i]);
        vector<int> right1;
        for(int i=flag1;i<postorder.size()-1;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,0,inorder.size()-1,postorder,0,postorder.size()-1);
    }
    
    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-1,postorder,begin2,begin2+leftlen-1);
        root->right=helper(inorder,begin1+leftlen+1,end1,postorder,begin2+leftlen,end2-1);
        return root;
    }
};