Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. http://www.leetcode.com/onlinejudge

leetcode_construct_binary_tree_from_preorder_and_inorder_traversal.cpp

#include <tr1::unordered_map>
using namespace std;

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

typedef tr1::unordered_map<int, int> hash_;

class Solution {
public:
    TreeNode* build_(hash_& h, vector<int>& pre, int pre_left, int pre_right, vector<int>& in, int in_left, int in_right) {
        if (pre_left > pre_right) {
            return NULL;
        }
        
        TreeNode* nd = new TreeNode( pre[pre_left] );
        if (pre_left == pre_right) {
            return nd;
        }
        
        int in_root_id = h[ pre[pre_left] ];
        int left_len = in_root_id - in_left;
        nd->left = build_(h, pre, pre_left + 1, pre_left + left_len, in, in_left, in_root_id - 1);
        nd->right = build_(h, pre, pre_left + left_len + 1, pre_right, in, in_root_id + 1, in_right);
        return nd;
    }

    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        if (inorder.empty()) {
            return NULL;
        }
        
        hash_ h;
        for (int i = 0; i < inorder.size(); ++i) {
            h[ inorder[i] ] = i;
        }
        
        return build_(h, preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
    }
};