What is the brute force solution for Construct Binary Tree from Inorder and Postorder Traversal?

The brute force approach for Construct Binary Tree from Inorder and Postorder Traversal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves repeatedly searching for the root node in the inorder array and constructing the tree recursively. This is straightforward but inefficient due to repeated searches. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Construct Binary Tree from Inorder and Postorder Traversal?

Construct Binary Tree from Inorder and Postorder Traversal uses the following concepts: Array, Hash Table, Divide and Conquer, Tree, Binary Tree. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Construct Binary Tree from Inorder and Postorder Traversal?

Explain your thought process clearly. Consider edge cases and how your solution handles them. Practice building trees from different traversal methods.

What are common mistakes when solving Construct Binary Tree from Inorder and Postorder Traversal?

Not handling the base case of empty arrays. Forgetting to update the bounds correctly during recursion.

#106

Construct Binary Tree from Inorder and Postorder Traversal

Medium

ArrayHash TableDivide and ConquerTreeBinary TreeHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a hashmap to store the indices of elements in the inorder array, allowing for O(1) access time. This reduces the overall time complexity significantly.

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Algorithm

4 steps

1Step 1: Create a hashmap to store the indices of elements in the inorder array.
2Step 2: Define a recursive function that takes the current bounds of the inorder and postorder arrays.
3Step 3: Identify the root from the postorder array and use the hashmap to find its index in the inorder array in O(1).
4Step 4: Recursively construct the right and left subtrees using the updated bounds.

solution.py19 lines

1class TreeNode:
2    def __init__(self, val=0, left=None, right=None):
3        self.val = val
4        self.left = left
5        self.right = right
6
7class Solution:
8    def buildTree(self, inorder, postorder):
9        index_map = {val: idx for idx, val in enumerate(inorder)}
10        def build(in_left, in_right):
11            if in_left > in_right:
12                return None
13            root_val = postorder.pop()
14            root = TreeNode(root_val)
15            index = index_map[root_val]
16            root.right = build(index + 1, in_right)
17            root.left = build(in_left, index - 1)
18            return root
19        return build(0, len(inorder) - 1)

ℹ

Complexity note: The time complexity is O(n) because we traverse the inorder and postorder arrays once to build the tree and create the hashmap. The space complexity is O(n) due to the hashmap storing indices.

1Understanding tree traversal methods is crucial.
2Using a hashmap can significantly reduce lookup times.

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