105. Construct Binary Tree from Preorder and Inorder Traversal 🔗

Difficulty: Medium - Tags: Binary Tree, Divide and Conquer, Recursion

LeetCode Problem Link

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Problem Statement 📜

Given two integer arrays preorder and inorder:

• preorder represents the preorder traversal of a binary tree.

• inorder represents the inorder traversal of the same binary tree.

Construct and return the binary tree.

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Examples 🌟

🔹 Example 1:

Input:

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Output:

[3,9,20,null,null,15,7]

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🔹 Example 2:

Input:

preorder = [-1]
inorder = [-1]

Output:

[-1]

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Constraints ⚙️

• 1 <= preorder.length <= 3000

• inorder.length == preorder.length

• -3000 <= preorder[i], inorder[i] <= 3000

• preorder and inorder consist of unique values.

• Each value in 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.

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Solution 💡

To construct the binary tree:

1. The first value in preorder is the root node.

2. Find the root node's position in inorder. Values to the left belong to the left subtree, and values to the right belong to the right subtree.

3. Recursively repeat this process for the left and right subtrees.

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Java Solution

import java.util.HashMap;

class Solution {
    private int preorderIndex = 0;
    private HashMap<Integer, Integer> inorderIndexMap;

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        // Create a map to store the index of each value in the inorder array
        inorderIndexMap = new HashMap<>();
        for (int i = 0; i < inorder.length; i++) {
            inorderIndexMap.put(inorder[i], i);
        }

        return buildSubtree(preorder, 0, inorder.length - 1);
    }

    private TreeNode buildSubtree(int[] preorder, int left, int right) {
        if (left > right) {
            return null; // Base case: no elements to construct the tree
        }

        // Get the current root value from preorder
        int rootValue = preorder[preorderIndex++];
        TreeNode root = new TreeNode(rootValue);

        // Find the index of the root value in the inorder array
        int inorderIndex = inorderIndexMap.get(rootValue);

        // Recursively construct the left and right subtrees
        root.left = buildSubtree(preorder, left, inorderIndex - 1);
        root.right = buildSubtree(preorder, inorderIndex + 1, right);

        return root;
    }
}

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Explanation of the Solution

1. Preorder Traversal:

   • The first element is the root.

   • Subsequent elements belong to the left or right subtree.

2. Inorder Traversal:

   • Left subtree elements come before the root.

   • Right subtree elements come after the root.

3. Recursive Construction:

   • Use the preorder array to pick the root.

   • Divide the inorder array into left and right subtrees.

   • Recursively construct the tree for both subtrees.

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Time Complexity ⏳

• O(n), where n is the number of nodes. Each node is visited once, and the HashMap provides O(1) lookups for the index.

Space Complexity 💾

• O(n) for the HashMap and recursive stack space.

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Follow-up Challenges 🧐

• What changes would you make to solve the problem if the tree was not guaranteed to have unique values?

• How would the approach differ for constructing a binary search tree?

You can find the full solution here.