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 all left leaves.
A leaf is a node with no children. A left leaf is a leaf that is the left child of another node.
Example 1:
Input: root = [3,9,20,null,null,15,7] Output: 24 Explanation: There are two left leaves in the binary tree, with values 9 and 15 respectively.
Example 2:
Input: root = [1] Output: 0
Constraints:
[1, 1000].-1000 <= Node.val <= 1000Problem summary: Given the root of a binary tree, return the sum of all left leaves. A leaf is a node with no children. A left leaf is a leaf that is the left child of another node.
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]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #404: Sum of Left Leaves
/**
* 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 int sumOfLeftLeaves(TreeNode root) {
if (root == null) {
return 0;
}
int ans = sumOfLeftLeaves(root.right);
if (root.left != null) {
if (root.left.left == root.left.right) {
ans += root.left.val;
} else {
ans += sumOfLeftLeaves(root.left);
}
}
return ans;
}
}
// Accepted solution for LeetCode #404: Sum of Left Leaves
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func sumOfLeftLeaves(root *TreeNode) int {
if root == nil {
return 0
}
ans := sumOfLeftLeaves(root.Right)
if root.Left != nil {
if root.Left.Left == root.Left.Right {
ans += root.Left.Val
} else {
ans += sumOfLeftLeaves(root.Left)
}
}
return ans
}
# Accepted solution for LeetCode #404: Sum of Left Leaves
# 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 sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:
if root is None:
return 0
ans = self.sumOfLeftLeaves(root.right)
if root.left:
if root.left.left == root.left.right:
ans += root.left.val
else:
ans += self.sumOfLeftLeaves(root.left)
return ans
// Accepted solution for LeetCode #404: Sum of Left Leaves
// 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 {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, is_left: bool) -> i32 {
if root.is_none() {
return 0;
}
let node = root.as_ref().unwrap().borrow();
let left = &node.left;
let right = &node.right;
if left.is_none() && right.is_none() {
if is_left {
return node.val;
}
return 0;
}
Self::dfs(left, true) + Self::dfs(right, false)
}
pub fn sum_of_left_leaves(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
Self::dfs(&root, false)
}
}
// Accepted solution for LeetCode #404: Sum of Left Leaves
/**
* 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 sumOfLeftLeaves(root: TreeNode | null): number {
if (!root) {
return 0;
}
let ans = sumOfLeftLeaves(root.right);
if (root.left) {
if (root.left.left === root.left.right) {
ans += root.left.val;
} else {
ans += sumOfLeftLeaves(root.left);
}
}
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.