leetcode——106——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 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(begin2 > end2)            return NULL;        else if(begin2 == end2)            return new TreeNode(postorder[end2]);                    TreeNode* root = new TreeNode(postorder[end2]);        int i = begin1;        for(; i <= end1; i ++)        {            if(inorder[i] == postorder[end2])                break;        }        //inorder[i] is the root        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;    }};