notes/OJ notes/pages/Leetcode Insert-Into-a-Binary-Search-Tree.md

Leetcode Insert-Into-a-Binary-Search-Tree

2022-07-07 08:50

Algorithms:

#algorithm #recursion #DFS

Data structures:

#DS #binary_tree

Difficulty:

#coding_problem #difficulty-medium

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

Links:

• Link to problem

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Problem

You are given the root node of a binary search tree (BST) and a value to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.

Notice that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.

Examples

Example 1:

Input: root = [4,2,7,1,3], val = 5 Output: [4,2,7,1,3,5] Explanation: Another accepted tree is:

Example 2:

Input: root = [40,20,60,10,30,50,70], val = 25 Output: [40,20,60,10,30,50,70,null,null,25]

Example 3:

Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5 Output: [4,2,7,1,3,5]

Constraints

• The number of nodes in the tree will be in the range [0, 104].
• -108 <= Node.val <= 108
• All the values Node.val are unique.
• -108 <= val <= 108
• It's guaranteed that val does not exist in the original BST.

Thoughts

[!summary] This is a #DFS problem.

Seems that no one seems to care about the second way, since it's way too complex.

DFS

DFS-like solution is a simple recursion problem, using a helper function.

Edge case:

root is null, simple return a new node.

Base cases:

• left side is empty, and val < root->val: place it in
• right side is empty, and val > root->val: place it in

Pseudocode:

• check for base cases
• if val < root->val, insert(root->left, val)
• vice, versa.

BFS

Solution