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.
Build confidence with an intuition-first walkthrough focused on tree fundamentals.
Given a binary tree, determine if it is height-balanced.
Example 1:
Input: root = [3,9,20,null,null,15,7] Output: true
Example 2:
Input: root = [1,2,2,3,3,null,null,4,4] Output: false
Example 3:
Input: root = [] Output: true
Constraints:
[0, 5000].-104 <= Node.val <= 104Problem summary: Given a binary tree, determine if it is height-balanced. Example 1: Input: root = [3,9,20,null,null,15,7] Output: true Example 2: Input: root = [1,2,2,3,3,null,null,4,4] Output: false Example 3: Input: root = [] Output: true
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree
[3,9,20,null,null,15,7]
[1,2,2,3,3,null,null,4,4]
[]
maximum-depth-of-binary-tree)k-th-largest-perfect-subtree-size-in-binary-tree)check-balanced-string)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #110: Balanced Binary 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 boolean isBalanced(TreeNode root) {
return height(root) >= 0;
}
private int height(TreeNode root) {
if (root == null) {
return 0;
}
int l = height(root.left);
int r = height(root.right);
if (l == -1 || r == -1 || Math.abs(l - r) > 1) {
return -1;
}
return 1 + Math.max(l, r);
}
}
// Accepted solution for LeetCode #110: Balanced Binary Tree
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func isBalanced(root *TreeNode) bool {
var height func(*TreeNode) int
height = func(root *TreeNode) int {
if root == nil {
return 0
}
l, r := height(root.Left), height(root.Right)
if l == -1 || r == -1 || abs(l-r) > 1 {
return -1
}
if l > r {
return 1 + l
}
return 1 + r
}
return height(root) >= 0
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
# Accepted solution for LeetCode #110: Balanced Binary 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 isBalanced(self, root: Optional[TreeNode]) -> bool:
def height(root):
if root is None:
return 0
l, r = height(root.left), height(root.right)
if l == -1 or r == -1 or abs(l - r) > 1:
return -1
return 1 + max(l, r)
return height(root) >= 0
// Accepted solution for LeetCode #110: Balanced Binary Tree
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
pub fn is_balanced(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
Self::dfs(&root) > -1
}
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
if root.is_none() {
return 0;
}
let node = root.as_ref().unwrap().borrow();
let left = Self::dfs(&node.left);
let right = Self::dfs(&node.right);
if left == -1 || right == -1 || (left - right).abs() > 1 {
return -1;
}
1 + left.max(right)
}
}
// Accepted solution for LeetCode #110: Balanced Binary 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 isBalanced(root: TreeNode | null): boolean {
const dfs = (root: TreeNode | null) => {
if (root == null) {
return 0;
}
const left = dfs(root.left);
const right = dfs(root.right);
if (left === -1 || right === -1 || Math.abs(left - right) > 1) {
return -1;
}
return 1 + Math.max(left, right);
};
return dfs(root) > -1;
}
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.