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 the root of a binary tree, return the sum of every tree node's tilt.
The tilt of a tree node is the absolute difference between the sum of all left subtree node values and all right subtree node values. If a node does not have a left child, then the sum of the left subtree node values is treated as 0. The rule is similar if the node does not have a right child.
Example 1:
Input: root = [1,2,3] Output: 1 Explanation: Tilt of node 2 : |0-0| = 0 (no children) Tilt of node 3 : |0-0| = 0 (no children) Tilt of node 1 : |2-3| = 1 (left subtree is just left child, so sum is 2; right subtree is just right child, so sum is 3) Sum of every tilt : 0 + 0 + 1 = 1
Example 2:
Input: root = [4,2,9,3,5,null,7] Output: 15 Explanation: Tilt of node 3 : |0-0| = 0 (no children) Tilt of node 5 : |0-0| = 0 (no children) Tilt of node 7 : |0-0| = 0 (no children) Tilt of node 2 : |3-5| = 2 (left subtree is just left child, so sum is 3; right subtree is just right child, so sum is 5) Tilt of node 9 : |0-7| = 7 (no left child, so sum is 0; right subtree is just right child, so sum is 7) Tilt of node 4 : |(3+5+2)-(9+7)| = |10-16| = 6 (left subtree values are 3, 5, and 2, which sums to 10; right subtree values are 9 and 7, which sums to 16) Sum of every tilt : 0 + 0 + 0 + 2 + 7 + 6 = 15
Example 3:
Input: root = [21,7,14,1,1,2,2,3,3] Output: 9
Constraints:
[0, 104].-1000 <= Node.val <= 1000Problem summary: Given the root of a binary tree, return the sum of every tree node's tilt. The tilt of a tree node is the absolute difference between the sum of all left subtree node values and all right subtree node values. If a node does not have a left child, then the sum of the left subtree node values is treated as 0. The rule is similar if the node does not have a right child.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree
[1,2,3]
[4,2,9,3,5,null,7]
[21,7,14,1,1,2,2,3,3]
find-all-the-lonely-nodes)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #563: Binary Tree Tilt
/**
* 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 {
private int ans;
public int findTilt(TreeNode root) {
dfs(root);
return ans;
}
private int dfs(TreeNode root) {
if (root == null) {
return 0;
}
int l = dfs(root.left), r = dfs(root.right);
ans += Math.abs(l - r);
return l + r + root.val;
}
}
// Accepted solution for LeetCode #563: Binary Tree Tilt
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func findTilt(root *TreeNode) (ans int) {
var dfs func(*TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
l, r := dfs(root.Left), dfs(root.Right)
ans += abs(l - r)
return l + r + root.Val
}
dfs(root)
return
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
# Accepted solution for LeetCode #563: Binary Tree Tilt
# 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 findTilt(self, root: Optional[TreeNode]) -> int:
def dfs(root: Optional[TreeNode]) -> int:
if root is None:
return 0
l, r = dfs(root.left), dfs(root.right)
nonlocal ans
ans += abs(l - r)
return l + r + root.val
ans = 0
dfs(root)
return ans
// Accepted solution for LeetCode #563: Binary Tree Tilt
struct Solution;
use rustgym_util::*;
trait Tilt {
fn find_tilt(&self, tilt: &mut i32) -> i32;
}
impl Tilt for TreeLink {
fn find_tilt(&self, tilt: &mut i32) -> i32 {
if let Some(node) = self {
let node = node.borrow();
let left = &node.left;
let right = &node.right;
let left_sum = left.find_tilt(tilt);
let right_sum = right.find_tilt(tilt);
*tilt += (left_sum - right_sum).abs();
node.val + left_sum + right_sum
} else {
0
}
}
}
impl Solution {
fn find_tilt(root: TreeLink) -> i32 {
let mut tilt = 0;
root.find_tilt(&mut tilt);
tilt
}
}
#[test]
fn test() {
let root = tree!(1, tree!(2), tree!(3));
assert_eq!(Solution::find_tilt(root), 1);
}
// Accepted solution for LeetCode #563: Binary Tree Tilt
/**
* 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 findTilt(root: TreeNode | null): number {
let ans: number = 0;
const dfs = (root: TreeNode | null): number => {
if (!root) {
return 0;
}
const [l, r] = [dfs(root.left), dfs(root.right)];
ans += Math.abs(l - r);
return l + r + root.val;
};
dfs(root);
return ans;
}
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.