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 roots of two binary trees root and subRoot, return true if there is a subtree of root with the same structure and node values of subRoot and false otherwise.
A subtree of a binary tree tree is a tree that consists of a node in tree and all of this node's descendants. The tree tree could also be considered as a subtree of itself.
Example 1:
Input: root = [3,4,5,1,2], subRoot = [4,1,2] Output: true
Example 2:
Input: root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2] Output: false
Constraints:
root tree is in the range [1, 2000].subRoot tree is in the range [1, 1000].-104 <= root.val <= 104-104 <= subRoot.val <= 104Problem summary: Given the roots of two binary trees root and subRoot, return true if there is a subtree of root with the same structure and node values of subRoot and false otherwise. A subtree of a binary tree tree is a tree that consists of a node in tree and all of this node's descendants. The tree tree could also be considered as a subtree of itself.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree · String Matching
[3,4,5,1,2] [4,1,2]
[3,4,5,1,2,null,null,null,null,0] [4,1,2]
count-univalue-subtrees)most-frequent-subtree-sum)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #572: Subtree of Another 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 isSubtree(TreeNode root, TreeNode subRoot) {
if (root == null) {
return false;
}
return same(root, subRoot) || isSubtree(root.left, subRoot)
|| isSubtree(root.right, subRoot);
}
private boolean same(TreeNode p, TreeNode q) {
if (p == null || q == null) {
return p == q;
}
return p.val == q.val && same(p.left, q.left) && same(p.right, q.right);
}
}
// Accepted solution for LeetCode #572: Subtree of Another Tree
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func isSubtree(root *TreeNode, subRoot *TreeNode) bool {
var same func(p, q *TreeNode) bool
same = func(p, q *TreeNode) bool {
if p == nil || q == nil {
return p == q
}
return p.Val == q.Val && same(p.Left, q.Left) && same(p.Right, q.Right)
}
if root == nil {
return false
}
return same(root, subRoot) || isSubtree(root.Left, subRoot) || isSubtree(root.Right, subRoot)
}
# Accepted solution for LeetCode #572: Subtree of Another 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 isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
def same(p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
if p is None or q is None:
return p is q
return p.val == q.val and same(p.left, q.left) and same(p.right, q.right)
if root is None:
return False
return (
same(root, subRoot)
or self.isSubtree(root.left, subRoot)
or self.isSubtree(root.right, subRoot)
)
// Accepted solution for LeetCode #572: Subtree of Another 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_subtree(
root: Option<Rc<RefCell<TreeNode>>>,
sub_root: Option<Rc<RefCell<TreeNode>>>,
) -> bool {
if root.is_none() {
return false;
}
Self::same(&root, &sub_root)
|| Self::is_subtree(
root.as_ref().unwrap().borrow().left.clone(),
sub_root.clone(),
)
|| Self::is_subtree(
root.as_ref().unwrap().borrow().right.clone(),
sub_root.clone(),
)
}
fn same(p: &Option<Rc<RefCell<TreeNode>>>, q: &Option<Rc<RefCell<TreeNode>>>) -> bool {
match (p, q) {
(None, None) => true,
(Some(p), Some(q)) => {
let p = p.borrow();
let q = q.borrow();
p.val == q.val && Self::same(&p.left, &q.left) && Self::same(&p.right, &q.right)
}
_ => false,
}
}
}
// Accepted solution for LeetCode #572: Subtree of Another 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 isSubtree(root: TreeNode | null, subRoot: TreeNode | null): boolean {
const same = (p: TreeNode | null, q: TreeNode | null): boolean => {
if (!p || !q) {
return p === q;
}
return p.val === q.val && same(p.left, q.left) && same(p.right, q.right);
};
if (!root) {
return false;
}
return same(root, subRoot) || isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot);
}
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.