LeetCode OJ--Construct Binary Tree from Inorder and Postorder Traversal *

http://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

知道二叉树的中序遍历和后序遍历序列，求二叉树。

使用递归

#include <iostream>
#include <vector>
using namespace std;

 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 buildTree(begin(inorder),end(inorder),begin(postorder),end(postorder));
    }
    template<typename BidiIt>
    TreeNode* buildTree(BidiIt in_first, BidiIt in_last,BidiIt post_first,BidiIt post_last)
    {
        if(in_first ==in_last) 
            return nullptr;
        if(post_first == post_last)
            return nullptr;
        const auto val = *prev(post_last);
        TreeNode* root = new TreeNode(val);

        auto in_root_pos = find(in_first,in_last,val);
        auto left_size = distance(in_first,in_root_pos);
        auto post_left_last = next(post_first,left_size);

        root->left = buildTree(in_first,in_root_pos,post_first,post_left_last);
        root->right = buildTree(next(in_root_pos),in_last,post_left_last,prev(post_last));
        return root;
    }
};

int main()
{
    Solution myS;
    int arr1[7] = {4,2};//,5,1,6,3,7 };
    int arr2[7] = {4,2};//,2,6,7,3,1 };
    vector<int> inorder(arr1,arr1+2) ;
    vector<int> postorder(arr2 ,arr2+2);
    TreeNode *myNode;

    myNode = myS.buildTree(inorder,postorder);
     
    cout<<"hi"<<endl;
    return 0;
}