5 Best Ways to Find the Largest Value in a Tree Using Inorder Traversal in Python

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πŸ’‘ Problem Formulation: In computer science, a common problem is determining the largest value in a binary tree via inorder traversal. Given a binary tree, the goal is to traverse its nodes in inorder sequence (left node, root node, right node) and locate the node with the maximum value. For instance, in a tree with values 1, 3, and 2 as its nodes, an inorder traversal would yield 1, 2, 3 and the desired output would be 3.

Method 1: Iterative Approach Using Stack

This method employs an iterative approach to traverse the tree using a stack. The iterative method tends to be faster than the recursive approach and consumes less memory on large trees. Multiple calls are made while processing each node, with the highest value tracked through a variable.

Here’s an example:

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def find_largest_inorder_iterative(root):
    stack, max_val = [], float('-inf')

    while stack or root:
        while root:
            root = root.left
        root = stack.pop()
        max_val = max(max_val, root.val)
        root = root.right
    return max_val

# Example usage:
root = TreeNode(1, None, TreeNode(3, TreeNode(2)))

Output: 3

This snippet iteratively performs an inorder traversal on the tree. A stack is used to keep track of yet-to-be-visited nodes. As we traverse, we update the max_val only when popping from the stack, ensuring that each node’s value is compared with the current maximum.

Method 2: Recursive Approach

The recursive approach to find the largest value in a binary tree involves a helper function that traverses the tree in preorder and returns the maximum value found. This is an elegant solution, but can lead to stack overflow with very deep trees.

Here’s an example:

def find_largest_inorder_recursive(root):
    def inorder_traverse(node):
        if not node:
            return float('-inf')
        left_max = inorder_traverse(node.left)
        right_max = inorder_traverse(node.right)
        return max(node.val, left_max, right_max)

    return inorder_traverse(root)

# Example usage:

Output: 3

In this code, inorder_traverse is a recursive function that checks if the current node is null, and if not, it compares the current node’s value with the max of its left and right children. The inorder_traverse function is then called with the root to start the traversal.

Method 3: Morris Traversal

Morris traversal is a space-efficient method that finds the largest value in a binary tree without using extra space for a stack or recursion call stack. It traverses the tree by temporarily modifying it, which might not be desirable with read-only or immutable tree structures.

Here’s an example:

def find_largest_inorder_morris(root):
    current, max_val = root, float('-inf')

    while current:
        if current.left is None:
            max_val = max(max_val, current.val)
            current = current.right
            pre = current.left
            while pre.right and pre.right != current:
                pre = pre.right

            if not pre.right:
                pre.right = current
                current = current.left
                pre.right = None
                max_val = max(max_val, current.val)
                current = current.right

    return max_val

# Example usage:

Output: 3

This code uses Morris traversal technique to traverse the tree. We use the left child’s rightmost node to keep track of nodes in current traversal, and modify it to temporarily point to the current node. It is later restored back as we find the largest value.

Method 4: Divide and Conquer

This strategy employs a divide and conquer paradigm, breaking the tree down into sub-trees and locating the largest value in each before combining the results. It is especially effective for balanced trees but is less useful on unbalanced ones.

Here’s an example:

def find_largest_inorder_divide_and_conquer(root):
    if root is None:
        return float('-inf')
    left_max = find_largest_inorder_divide_and_conquer(root.left)
    right_max = find_largest_inorder_divide_and_conquer(root.right)
    return max(root.val, left_max, right_max)

# Example usage:

Output: 3

The snippet implements a divide and conquer approach, where the tree is split into left and right sub-trees. The function is called recursively on each side, and the largest values from each sub-tree are compared along with the current node’s value to determine the maximum value in the tree.

Bonus One-Liner Method 5: Using Generator Expression with Inorder Traversal

This one-liner leverages generators for a compact and elegant solution. It traverses the tree in inorder sequence using a generator expression and finds the largest value. This method is concise but may not be as readable for beginners.

Here’s an example:

def find_largest_inorder_generator(root):
    inorder = (node.val for node in iter_tree(root))
    return max(inorder)

def iter_tree(node):
    if node:
        yield from iter_tree(node.left)
        yield node
        yield from iter_tree(node.right)

# Example usage:

Output: 3

This code leverages Python’s generator expression within the max function to perform the inorder traversal. It uses a separate generator function iter_tree that yields each node as it traverses the tree inorder.


  • Method 1: Iterative Approach Using Stack. Fast and memory efficient. Less intuitive for those familiar with recursion.
  • Method 2: Recursive Approach. Simple and elegant. Potentially causes stack overflow on very deep trees.
  • Method 3: Morris Traversal. Space-efficient, avoids stack or recursion. Temporarily modifies the tree.
  • Method 4: Divide and Conquer. Good for balanced trees. Less efficient for unbalanced trees.
  • Method 5: Using Generator Expression. Elegant and Pythonic. Less readable for those not familiar with generators.