106. Construct Binary Tree from Inorder and Postorder Traversal

Medium LeetCode

Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.

Example 1

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

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

Example 2

Input: inorder = [-1], postorder = [-1]

Output: [-1]

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.

How to solve the problem

Code

• Recursion (DFS)

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
        if not inorder or not postorder:
            return None
        
        # The last element in postorder is the root of the current subtree
        root_val = postorder.pop()
        root = TreeNode(root_val)

        # Find the index of the root in inorder to split left and right subtrees
        index = inorder.index(root_val)

        # Important: build right subtree first because we are popping from the end of postorder
        root.right = self.buildTree(inorder[index + 1:], postorder)
        root.left = self.buildTree(inorder[:index], postorder)
        
        return root

Complexity

• Time complexity: O(n), n == number of nodes
• Space complexity: O(n)