LeetCode-in-Cpp

105. Construct Binary Tree from Preorder and Inorder Traversal

Medium

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.

Solution

#include <vector>
#include <unordered_map>

/**
 * 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 j;
    std::unordered_map<int, int> map;

public:
    Solution() : j(0) {}

    TreeNode* buildTree(std::vector<int>& preorder, std::vector<int>& inorder) {
        map.clear();
        j = 0;
        for (int i = 0; i < inorder.size(); ++i) {
            map[inorder[i]] = i;
        }
        return answer(preorder, inorder, 0, preorder.size() - 1);
    }

private:
    TreeNode* answer(const std::vector<int>& preorder, const std::vector<int>& inorder, int start, int end) {
        if (start > end || j >= preorder.size()) {
            return nullptr;
        }
        int value = preorder[j++];
        int index = map[value];
        TreeNode* node = new TreeNode(value);
        node->left = answer(preorder, inorder, start, index - 1);
        node->right = answer(preorder, inorder, index + 1, end);
        return node;
    }
};

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