106. Construct Binary Tree from Inorder and Postorder Traversal 🔗

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

LeetCode Problem Link

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Problem Statement 📜

Given two integer arrays inorder and postorder:

• inorder represents the inorder traversal of a binary tree.

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

Construct and return the binary tree.

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Examples 🌟

🔹 Example 1:

Input:

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

Output:

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

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🔹 Example 2:

Input:

inorder = [-1]
postorder = [-1]

Output:

[-1]

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Constraints ⚙️

• 1 <= inorder.length <= 3000

• postorder.length == inorder.length

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

• inorder and postorder consist of unique values.

• Each value of postorder also appears in inorder.

• inorder is guaranteed to be the inorder traversal of the tree.

• postorder is guaranteed to be the postorder traversal of the tree.

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Solution 💡

To construct the binary tree:

1. The last value in postorder 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 postorderIndex;
    private HashMap<Integer, Integer> inorderIndexMap;

    public TreeNode buildTree(int[] inorder, int[] postorder) {
        // Initialize postorder index to the last element
        postorderIndex = postorder.length - 1;

        // 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(postorder, 0, inorder.length - 1);
    }

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

        // Get the current root value from postorder
        int rootValue = postorder[postorderIndex--];
        TreeNode root = new TreeNode(rootValue);

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

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

        return root;
    }
}

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Explanation of the Solution

1. Postorder Traversal:

   • The last element is the root.

   • The second-to-last element is part of the right subtree.

2. Inorder Traversal:

   • Left subtree elements come before the root.

   • Right subtree elements come after the root.

3. Recursive Construction:

   • Use the postorder 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 🧐

• Could you modify the solution if the tree had duplicate values?

• What if the input arrays were very large? How would you optimize memory usage?

You can find the full solution here.