5.2.1 Construct Binary Tree from Preorder and Inorder Traversal

/*
	Author: Annie Kim, [email protected]
	Date: May 16, 2013
	Problem: Construct Binary Tree from Preorder and Inorder Traversal
	Difficulty: Easy
	Source: http://leetcode.com/onlinejudge#question_105
	Notes:
	Given preorder and inorder traversal of a tree, construct the binary tree.
	Note:
	You may assume that duplicates do not exist in the tree.
	Solution: Recursion.
	*/
	/**
	* 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> &preorder, vector<int> &inorder) {
        return buildTreeRe(preorder.begin(), inorder.begin(), preorder.size());
    }

    TreeNode *buildTreeRe(vector<int>::iterator preorder, vector<int>::iterator inorder, int N) {
        if (N <= 0) return NULL;
        vector<int>::iterator it = find(inorder, inorder + N, *preorder);
        int pos = it - inorder;
        TreeNode *root = new TreeNode(*preorder);
        root->left = buildTreeRe(preorder+1, inorder, pos);
        root->right = buildTreeRe(preorder+1+pos, inorder+pos+1, N-1-pos);
        return root;
    }
};