Given inorder and postorder 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_tre_from_inorder_and_postorder_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) {}
 * };
 */
class Solution {
public:
    TreeNode *build_(tr1::unordered_map<int, int>& num2id, vector<int>& in, int in_left, int in_right, vector<int>& post, int post_left, int post_right) {
        if (in_left > in_right) {
            return NULL;
        }
        TreeNode* node = new TreeNode(post[post_right]);
        if (in_left == in_right) {
            return node;
        }
        int in_root_id = num2id[post[post_right]];
        int left_len = in_root_id - in_left;
        node->left = build_(num2id, in, in_left, in_root_id - 1, post, post_left, post_left + left_len - 1);
        node->right = build_(num2id, in, in_root_id + 1, in_right, post, post_left + left_len, post_right - 1);
        return node;
    }

    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        if (inorder.empty()) {
            return NULL;
        }
        
        tr1::unordered_map<int, int> num2id;
        for (int i = 0; i < inorder.size(); ++i) {
            num2id[inorder[i]] = i;
        }
        
        return build_(num2id, inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);
    }
};