LeetCode #872 — EASY

Leaf-Similar Trees

Build confidence with an intuition-first walkthrough focused on tree fundamentals.

The Problem

Problem Statement

Consider all the leaves of a binary tree, from left to right order, the values of those leaves form a leaf value sequence.

For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8).

Two binary trees are considered leaf-similar if their leaf value sequence is the same.

Return true if and only if the two given trees with head nodes root1 and root2 are leaf-similar.

Example 1:

Input: root1 = [3,5,1,6,2,9,8,null,null,7,4], root2 = [3,5,1,6,7,4,2,null,null,null,null,null,null,9,8]
Output: true

Example 2:

Input: root1 = [1,2,3], root2 = [1,3,2]
Output: false

Constraints:

The number of nodes in each tree will be in the range [1, 200].
Both of the given trees will have values in the range [0, 200].

Step 01

Brute Force Baseline

Problem summary: Consider all the leaves of a binary tree, from left to right order, the values of those leaves form a leaf value sequence. For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8). Two binary trees are considered leaf-similar if their leaf value sequence is the same. Return true if and only if the two given trees with head nodes root1 and root2 are leaf-similar.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

[3,5,1,6,2,9,8,null,null,7,4]
[3,5,1,6,7,4,2,null,null,null,null,null,null,9,8]

Example 2

[1,2,3]
[1,3,2]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #872: Leaf-Similar Trees
/**
 * 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 leafSimilar(TreeNode root1, TreeNode root2) {
        List<Integer> l1 = new ArrayList<>();
        List<Integer> l2 = new ArrayList<>();
        dfs(root1, l1);
        dfs(root2, l2);
        return l1.equals(l2);
    }

    private void dfs(TreeNode root, List<Integer> nums) {
        if (root.left == root.right) {
            nums.add(root.val);
            return;
        }
        if (root.left != null) {
            dfs(root.left, nums);
        }
        if (root.right != null) {
            dfs(root.right, nums);
        }
    }
}

// Accepted solution for LeetCode #872: Leaf-Similar Trees
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func leafSimilar(root1 *TreeNode, root2 *TreeNode) bool {
	l1, l2 := []int{}, []int{}
	var dfs func(*TreeNode, *[]int)
	dfs = func(root *TreeNode, nums *[]int) {
		if root.Left == root.Right {
			*nums = append(*nums, root.Val)
			return
		}
		if root.Left != nil {
			dfs(root.Left, nums)
		}
		if root.Right != nil {
			dfs(root.Right, nums)
		}
	}
	dfs(root1, &l1)
	dfs(root2, &l2)
	return reflect.DeepEqual(l1, l2)
}

# Accepted solution for LeetCode #872: Leaf-Similar Trees
# 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 leafSimilar(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
        def dfs(root: Optional[TreeNode], nums: List[int]) -> None:
            if root.left == root.right:
                nums.append(root.val)
                return
            if root.left:
                dfs(root.left, nums)
            if root.right:
                dfs(root.right, nums)

        l1, l2 = [], []
        dfs(root1, l1)
        dfs(root2, l2)
        return l1 == l2

// Accepted solution for LeetCode #872: Leaf-Similar Trees
// 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 leaf_similar(
        root1: Option<Rc<RefCell<TreeNode>>>,
        root2: Option<Rc<RefCell<TreeNode>>>,
    ) -> bool {
        let mut l1 = Vec::new();
        let mut l2 = Vec::new();
        Self::dfs(&root1, &mut l1);
        Self::dfs(&root2, &mut l2);
        l1 == l2
    }

    fn dfs(node: &Option<Rc<RefCell<TreeNode>>>, nums: &mut Vec<i32>) {
        if let Some(n) = node {
            let n = n.borrow();
            if n.left.is_none() && n.right.is_none() {
                nums.push(n.val);
                return;
            }
            if n.left.is_some() {
                Self::dfs(&n.left, nums);
            }
            if n.right.is_some() {
                Self::dfs(&n.right, nums);
            }
        }
    }
}

// Accepted solution for LeetCode #872: Leaf-Similar Trees
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #872: Leaf-Similar Trees
// /**
//  * 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 leafSimilar(TreeNode root1, TreeNode root2) {
//         List<Integer> l1 = new ArrayList<>();
//         List<Integer> l2 = new ArrayList<>();
//         dfs(root1, l1);
//         dfs(root2, l2);
//         return l1.equals(l2);
//     }
// 
//     private void dfs(TreeNode root, List<Integer> nums) {
//         if (root.left == root.right) {
//             nums.add(root.val);
//             return;
//         }
//         if (root.left != null) {
//             dfs(root.left, nums);
//         }
//         if (root.right != null) {
//             dfs(root.right, nums);
//         }
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

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.

DFS TRAVERSAL

O(n) time

O(h) space

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.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.