BST Property and Why It Matters in DSA Go - Execution Trace

Concept Flow - BST Property and Why It Matters

Start with root node

↓

Check new value vs current node

↓

Go left

↓

Repeat until leaf

↓

Insert new node here

The BST property guides where to place new values by comparing with nodes, going left if smaller, right if larger or equal, until a leaf is found.

Execution Sample

DSA Go

Insert(10)
Insert(5)
Insert(15)
Insert(7)

Insert values into BST following the property: left subtree has smaller values, right subtree has larger or equal values.

Execution Table

Step	Operation	Nodes in Tree	Pointer Changes	Visual State
1	Insert 10 as root	10	root -> 10	10 -> null
2	Insert 5: 5 < 10, go left	10 -> 5	10.left -> 5	10 / 5 -> null
3	Insert 15: 15 >= 10, go right	10 -> 5, 15	10.right -> 15	10 / \ 5 15 -> null
4	Insert 7: 7 < 10, go left; 7 >= 5, go right	10 -> 5 -> 7, 15	5.right -> 7	10 / \ 5 15 \ 7 -> null
5	End of insertions	10 -> 5 -> 7, 15	No pointer changes	10 / \ 5 15 \ 7 -> null

💡 All values inserted following BST property; no more insertions.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	Final
root	null	10	10	10	10	10
current	null	10	10	10	5	null
10.left	null	null	5	5	5	5
10.right	null	null	null	15	15	15
5.right	null	null	null	null	7	7

Key Moments - 3 Insights

Why do we go left when the new value is smaller than the current node?

Because the BST property requires all smaller values to be in the left subtree, as shown in execution_table step 2 where 5 < 10 leads to left insertion.

Why do we go right when the new value is equal or larger than the current node?

The BST property places equal or larger values in the right subtree, ensuring order. See step 3 where 15 >= 10 goes right.

What happens if we reach a null pointer during traversal?

We insert the new node there. In step 4, after going left and then right, 7 is inserted where 5.right was null.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, which pointer changes to insert 15?

A 10.left

B 10.right

C 5.left

D 5.right

💡 Hint

Check the 'Pointer Changes' column at step 3 in execution_table.

At which step does the tree first have a node with two children?

A Step 3

B Step 2

C Step 4

D Step 5

💡 Hint

Look at the 'Visual State' column and see when root 10 has both left and right children.

If we insert a value 5 again after step 4, where will it be placed?

A Left of 5

B Left of 7

C Right of 5

D Right of 7

💡 Hint

Recall BST property: equal or larger values go right; see step 4 traversal for 7.

Concept Snapshot

BST Property:
- Left subtree nodes < current node
- Right subtree nodes >= current node
Insertion:
- Compare new value with current node
- Go left if smaller, right if larger or equal
- Insert at null pointer
Maintains sorted order for fast search

Full Transcript

The BST property means every node's left subtree has smaller values and right subtree has larger or equal values. When inserting, we start at the root and compare the new value. If smaller, we go left; if larger or equal, we go right. We repeat until we find a null spot to insert the new node. This keeps the tree ordered for efficient searching. For example, inserting 10 creates the root. Inserting 5 goes left of 10. Inserting 15 goes right of 10. Inserting 7 goes right of 5. This process ensures the BST property is always maintained.