105. Construct Binary Tree from Preorder and Inorder Traversal

/*
 * @lc app=leetcode id=105 lang=cpp
 *
 * [105] Construct Binary Tree from Preorder and Inorder Traversal
 *
 * https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/
 *
 * algorithms
 * Medium (65.35%)
 * Likes:    15764
 * Dislikes: 569
 * Total Accepted:    1.5M
 * Total Submissions: 2.3M
 * Testcase Example:  '[3,9,20,15,7]\n[9,3,15,20,7]'
 *
 * Given two integer arrays preorder and inorder where preorder is the preorder
 * traversal of a binary tree and inorder is the inorder traversal of the same
 * tree, construct and return the binary tree.
 *
 *
 * Example 1:
 *
 *
 * Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
 * Output: [3,9,20,null,null,15,7]
 *
 *
 * Example 2:
 *
 *
 * Input: preorder = [-1], inorder = [-1]
 * Output: [-1]
 *
 *
 *
 * Constraints:
 *
 *
 * 1 <= preorder.length <= 3000
 * inorder.length == preorder.length
 * -3000 <= preorder[i], inorder[i] <= 3000
 * preorder and inorder consist of unique values.
 * Each value of inorder also appears in preorder.
 * preorder is guaranteed to be the preorder traversal of the tree.
 * inorder is guaranteed to be the inorder traversal of the tree.
 *
 *
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    int rootIndex = 0;
    unordered_map<int, int> indexMap;
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder)
    {
        int n = preorder.size();

        for(int i = 0; i < n; ++i)
        {
            indexMap[inorder[i]] = i;
        }
        return build(preorder, 0, n - 1);
    }

    TreeNode* build(vector<int>& preorder, int left, int right)
    {
        if(left > right) return nullptr;

        int rootValue = preorder[rootIndex++];

        TreeNode* root = new TreeNode(rootValue);

        root->left = build(preorder, left, indexMap[rootValue] - 1);

        root->right = build(preorder, indexMap[rootValue] + 1, right);
        return root;
    }
};
// @lc code=end

• T: $O(N)$
• S: $O(N)$