1.9 KiB
1.9 KiB
To insert a node into a binary search tree (BST) while keeping the tree balanced, we need to perform the following steps:
- Insert the node in the appropriate place according to the BST properties (left child < parent < right child).
- After insertion, rebalance the tree (if necessary) to ensure that it remains balanced. A tree is considered balanced if the height difference between the left and right subtrees of any node is at most 1 (this is the balanced binary tree or AVL tree property).
Given Binary Search Tree:
10
/ \
5 15
/ / \
2 12 20
Step 1: Insert 4 into the BST
We start by inserting 4 into the BST:
- 4 is less than 10, so we move to the left of 10.
- 4 is less than 5, so we move to the left of 5.
- 4 is greater than 2, so 4 will be placed as the right child of 2.
After insertion, the tree looks like this:
10
/ \
5 15
/ / \
2 12 20
\
4
Step 2: Balance the Tree (If Needed)
At this point, the tree is not balanced. Let's check the balance factors of the nodes:
- Node 2 has a left subtree height of 0 and a right subtree height of 1. Its balance factor is
0 - 1 = -1, which is acceptable. - Node 5 has a left subtree height of 1 and a right subtree height of 1. Its balance factor is
1 - 1 = 0, which is balanced. - Node 10 has a left subtree height of 2 (because of nodes 5, 2, and 4) and a right subtree height of 2 (because of nodes 15, 12, and 20). Its balance factor is
2 - 2 = 0, which is balanced.
Since the tree is balanced at all levels, no further rebalancing is necessary.
Final Balanced Tree:
The final tree after inserting 4 is:
10
/ \
5 15
/ / \
2 12 20
\
4
Conclusion:
After inserting 4, the tree remains balanced and no rotations are needed.