DSA · Theory

Find if there is a triplet in a balanced bst that adds to zero

Theory

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Find if there is a triplet in a balanced bst that adds to zero

Find if there is a triplet in a balanced bst that adds to zero is a DSA idea you prove with folder on disk, not with a slogan.

Find if there is a triplet in a balanced bst that adds to zero on the board: left < root < right. Then say what folder on disk looks like after.

Find if there is a triplet in a balanced bst that adds to zero exists so Kabir can keep folder on disk correct in a file explorer.

Use Find if there is a triplet in a balanced bst that adds to zero when folder on disk must stay clear. If a simpler DSA step works, use that instead.

If folder on disk breaks under Find if there is a triplet in a balanced bst that adds to zero, check unbalanced BST becoming a list first.

After Find if there is a triplet in a balanced bst that adds to zero, Kabir should still remember unbalanced BST becoming a list.

Close Find if there is a triplet in a balanced bst that adds to zero with: “If I skip it, folder on disk goes wrong like this: unbalanced BST becoming a list.”

Diagram

Exam tip

Board: folder on disk. Dry-run Find if there is a triplet in a balanced bst that adds to zero. Name the trap: unbalanced BST becoming a list.

Example

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

def insert(root, val):
    if root is None:
        return Node(val)
    if val < root.val:
        root.left = insert(root.left, val)
    elif val > root.val:
        root.right = insert(root.right, val)
    return root

root = None
for v in [5, 2, 7, 1]:
    root = insert(root, v)
print(root.val, root.left.val, root.right.val)

Find if there is a triplet in a balanced bst that adds to zero — bST invariant: left < node < right — enables log-time search when balanced.

Short notes

DefFind if there is a triplet in a balanced bst that adds to zero — Kabir uses it for folder on disk in a file explorer.
RuleFind if there is a triplet in a balanced bst that adds to zero → left < root < right.
RememberFind if there is a triplet in a balanced bst that adds to zero + time vs memory (a file explorer).
UseFind if there is a triplet in a balanced bst that adds to zero in a file explorer (folder on disk).
TrapFind if there is a triplet in a balanced bst that adds to zero — unbalanced BST becoming a list.
ExFind if there is a triplet in a balanced bst that adds to zero → folder on disk.

class Node: def __init__(self, val): self.val = val self.left = None self.right = None def insert(root, val): if root is None: return Node(val) if val < root.val: root.left = insert(root.left, val) elif val > root.val: root.right = insert(root.right, val) return root root = None for v in [5, 2, 7, 1]: root = insert(root, v) print(root.val, root.left.val, root.right.val)

Find if there is a triplet in a balanced bst that adds to zero

Example

Short notes

Questions

Find if there is a triplet in a balanced bst that adds to zero

Example

Short notes

Questions