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.
Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X.
Return the number of good nodes in the binary tree.
Example 1:
Input: root = [3,1,4,3,null,1,5] Output: 4 Explanation: Nodes in blue are good. Root Node (3) is always a good node. Node 4 -> (3,4) is the maximum value in the path starting from the root. Node 5 -> (3,4,5) is the maximum value in the path Node 3 -> (3,1,3) is the maximum value in the path.
Example 2:
Input: root = [3,3,null,4,2] Output: 3 Explanation: Node 2 -> (3, 3, 2) is not good, because "3" is higher than it.
Example 3:
Input: root = [1] Output: 1 Explanation: Root is considered as good.
Constraints:
[1, 10^5].[-10^4, 10^4].Problem summary: Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X. Return the number of good nodes in the binary tree.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree
[3,1,4,3,null,1,5]
[3,3,null,4,2]
[1]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1448: Count Good Nodes in 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 {
private int ans = 0;
public int goodNodes(TreeNode root) {
dfs(root, -100000);
return ans;
}
private void dfs(TreeNode root, int mx) {
if (root == null) {
return;
}
if (mx <= root.val) {
++ans;
mx = root.val;
}
dfs(root.left, mx);
dfs(root.right, mx);
}
}
// Accepted solution for LeetCode #1448: Count Good Nodes in Binary Tree
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func goodNodes(root *TreeNode) (ans int) {
var dfs func(*TreeNode, int)
dfs = func(root *TreeNode, mx int) {
if root == nil {
return
}
if mx <= root.Val {
ans++
mx = root.Val
}
dfs(root.Left, mx)
dfs(root.Right, mx)
}
dfs(root, -10001)
return
}
# Accepted solution for LeetCode #1448: Count Good Nodes in 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 goodNodes(self, root: TreeNode) -> int:
def dfs(root: TreeNode, mx: int):
if root is None:
return
nonlocal ans
if mx <= root.val:
ans += 1
mx = root.val
dfs(root.left, mx)
dfs(root.right, mx)
ans = 0
dfs(root, -1000000)
return ans
// Accepted solution for LeetCode #1448: Count Good Nodes in Binary Tree
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
pub fn good_nodes(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
let root = root.unwrap();
let val = root.borrow().val;
let mut count = 0;
let mut stack = Vec::with_capacity(10_000);
stack.push((root, val as i16));
while let Some((curr, mut max)) = stack.pop() {
let mut curr = curr.borrow_mut();
if curr.val as i16 >= max {
count += 1;
max = curr.val as i16;
}
if curr.left.is_some() {
let mut left = None;
std::mem::swap(&mut left, &mut curr.left);
stack.push((left.unwrap(), max));
}
if curr.right.is_some() {
let mut right = None;
std::mem::swap(&mut right, &mut curr.right);
stack.push((right.unwrap(), max));
}
}
count
}
}
// Accepted solution for LeetCode #1448: Count Good Nodes in 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 goodNodes(root: TreeNode | null): number {
let ans = 0;
const dfs = (root: TreeNode | null, mx: number) => {
if (!root) {
return;
}
if (mx <= root.val) {
++ans;
mx = root.val;
}
dfs(root.left, mx);
dfs(root.right, mx);
};
dfs(root, -1e6);
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.