Forgetting null/base-case handling
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.
Move from brute-force thinking to an efficient approach using tree strategy.
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.
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:
[0, 104].-108 <= Node.val <= 108Node.val are unique.-108 <= val <= 108val does not exist in the original BST.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree
[4,2,7,1,3] 5
[40,20,60,10,30,50,70] 25
[4,2,7,1,3,null,null,null,null,null,null] 5
search-in-a-binary-search-tree)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #701: Insert into a Binary Search Tree
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode insertIntoBST(TreeNode root, int val) {
if (root == null) {
return new TreeNode(val);
}
if (root.val > val) {
root.left = insertIntoBST(root.left, val);
} else {
root.right = insertIntoBST(root.right, val);
}
return root;
}
}
// Accepted solution for LeetCode #701: Insert into a Binary Search Tree
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func insertIntoBST(root *TreeNode, val int) *TreeNode {
if root == nil {
return &TreeNode{Val: val}
}
if root.Val > val {
root.Left = insertIntoBST(root.Left, val)
} else {
root.Right = insertIntoBST(root.Right, val)
}
return root
}
# Accepted solution for LeetCode #701: Insert into a Binary Search Tree
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if root is None:
return TreeNode(val)
if root.val > val:
root.left = self.insertIntoBST(root.left, val)
else:
root.right = self.insertIntoBST(root.right, val)
return root
// Accepted solution for LeetCode #701: Insert into a Binary Search Tree
struct Solution;
use rustgym_util::*;
trait Postorder {
fn insert(self, val: i32) -> Self;
}
impl Postorder for TreeLink {
fn insert(self, val: i32) -> Self {
if let Some(node) = self {
let node_val = node.borrow().val;
let left = node.borrow_mut().left.take();
let right = node.borrow_mut().right.take();
if node_val < val {
tree!(node_val, left, right.insert(val))
} else {
tree!(node_val, left.insert(val), right)
}
} else {
tree!(val)
}
}
}
impl Solution {
fn insert_into_bst(root: TreeLink, val: i32) -> TreeLink {
root.insert(val)
}
}
#[test]
fn test() {
let root = tree!(4, tree!(2, tree!(1), tree!(3)), tree!(7));
let val = 5;
let res = tree!(4, tree!(2, tree!(1), tree!(3)), tree!(7, tree!(5), None));
assert_eq!(Solution::insert_into_bst(root, val), res);
}
// Accepted solution for LeetCode #701: Insert into a Binary Search Tree
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function insertIntoBST(root: TreeNode | null, val: number): TreeNode | null {
if (!root) {
return new TreeNode(val);
}
if (root.val > val) {
root.left = insertIntoBST(root.left, val);
} else {
root.right = insertIntoBST(root.right, val);
}
return root;
}
Use this to step through a reusable interview workflow for this problem.
BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.
Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.
Review these before coding to avoid predictable interview regressions.
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.