Introduction

In this blog post, we will explore how to solve the LeetCode problem "1325" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a binary tree root and an integer target, delete all the leaf nodes with value target. Note that once you delete a leaf node with value target, if its parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you cannot). Example 1: Input: root = [1,2,3,2,null,2,4], target = 2 Output: [1,null,3,null,4] Explanation: Leaf nodes in green with value (target = 2) are removed (Picture in left). After removing, new nodes become leaf nodes with value (target = 2) (Picture in center). Example 2: Input: root = [1,3,3,3,2], target = 3 Output: [1,3,null,null,2] Example 3: Input: root = [1,2,null,2,null,2], target = 2 Output: [1] Explanation: Leaf nodes in green with value (target = 2) are removed at each step. Constraints: The number of nodes in the tree is in the range [1, 3000]. 1 <= Node.val, target <= 1000

Explanation

Here's a solution to the problem of deleting leaf nodes with a specific target value in a binary tree, along with explanations:

• Bottom-Up Traversal: Use a post-order (left, right, root) traversal of the binary tree. This ensures that we process the children of a node before processing the node itself.
• Leaf Node Check: At each node, check if it's a leaf node and if its value equals the target. If both conditions are true, remove the node by returning None.

• Recursive Deletion: Recursively apply the same logic to the left and right children of each node. If a child becomes None after deletion (because it was a target leaf), update the parent's corresponding child pointer to None.

• Time Complexity: O(N), where N is the number of nodes in the tree. We visit each node once. Space Complexity: O(H), where H is the height of the tree, due to the recursion stack. In the worst case (skewed tree), H = N, but in a balanced tree, H = log N.

Code

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

def remove_leaf_nodes(root, target):
    """
    Removes all leaf nodes with value target from a binary tree.

    Args:
        root: The root of the binary tree.
        target: The target value to remove.

    Returns:
        The root of the modified binary tree.
    """

    def remove_leaves(node, target):
        if not node:
            return None

        node.left = remove_leaves(node.left, target)
        node.right = remove_leaves(node.right, target)

        if not node.left and not node.right and node.val == target:
            return None

        return node

    return remove_leaves(root, target)