DSA · Theory

Count Non-Leaf Nodes in a Binary Tree

Theory

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Count Non-Leaf Nodes in a Binary Tree

If you can teach Count Non-Leaf Nodes in a Binary Tree (alt) using folder on disk, you know it. If you only know the heading, you don’t.

Count Non-Leaf Nodes in a Binary Tree (alt) on the board: left < root < right. Then say what folder on disk looks like after.

Skip Count Non-Leaf Nodes in a Binary Tree (alt) and unbalanced BST becoming a list shows up in a file explorer.

Ananya ships Count Non-Leaf Nodes in a Binary Tree (alt) in a file explorer. That is the use case worth saying.

Count Non-Leaf Nodes in a Binary Tree (alt) miss: unbalanced BST becoming a list. Fix it before you talk about advanced DSA.

After Count Non-Leaf Nodes in a Binary Tree (alt), Ananya should still remember unbalanced BST becoming a list.

Close Count Non-Leaf Nodes in a Binary Tree (alt) with: “If I skip it, folder on disk goes wrong like this: unbalanced BST becoming a list.”

Exam tip

Say Count Non-Leaf Nodes in a Binary Tree (alt) in one breath, then left < root < right, then unbalanced BST becoming a list.

Example

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

def inorder(n):
    if not n:
        return []
    return inorder(n.left) + [n.val] + inorder(n.right)

root = Node(2, Node(1), Node(3))
print(inorder(root))

Count Non-Leaf Nodes in a Binary Tree (alt) — inorder walks left → node → right (sorted order for a BST).

Short notes

DefCount Non-Leaf Nodes in a Binary Tree (alt) — Ananya uses it for folder on disk in a file explorer.
RuleCount Non-Leaf Nodes in a Binary Tree (alt) → left < root < right.
RememberCount Non-Leaf Nodes in a Binary Tree (alt) + time vs memory (a file explorer).
UseCount Non-Leaf Nodes in a Binary Tree (alt) in a file explorer (folder on disk).
TrapCount Non-Leaf Nodes in a Binary Tree (alt) — unbalanced BST becoming a list.
ExCount Non-Leaf Nodes in a Binary Tree (alt) → folder on disk.

class Node: def __init__(self, val, left=None, right=None): self.val, self.left, self.right = val, left, right def inorder(n): if not n: return [] return inorder(n.left) + [n.val] + inorder(n.right) root = Node(2, Node(1), Node(3)) print(inorder(root))

Count Non-Leaf Nodes in a Binary Tree

Example

Short notes

Questions

Count Non-Leaf Nodes in a Binary Tree

Example

Short notes

Questions