LeetCode #951 — MEDIUM

Flip Equivalent Binary Trees

Move from brute-force thinking to an efficient approach using tree strategy.

The Problem

Problem Statement

For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.

A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.

Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivalent or false otherwise.

Example 1:

Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
Output: true
Explanation: We flipped at nodes with values 1, 3, and 5.

Example 2:

Input: root1 = [], root2 = []
Output: true

Example 3:

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

Constraints:

The number of nodes in each tree is in the range [0, 100].
Each tree will have unique node values in the range [0, 99].

Step 01

Brute Force Baseline

Problem summary: For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees. A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations. Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivalent or false otherwise.

Baseline thinking

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

Pattern signal: Tree

Example 1

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

Example 2

[]
[]

Example 3

[]
[1]

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 #951: Flip Equivalent Binary 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 flipEquiv(TreeNode root1, TreeNode root2) {
        return dfs(root1, root2);
    }

    private boolean dfs(TreeNode root1, TreeNode root2) {
        if (root1 == root2 || (root1 == null && root2 == null)) {
            return true;
        }
        if (root1 == null || root2 == null || root1.val != root2.val) {
            return false;
        }
        return (dfs(root1.left, root2.left) && dfs(root1.right, root2.right))
            || (dfs(root1.left, root2.right) && dfs(root1.right, root2.left));
    }
}

// Accepted solution for LeetCode #951: Flip Equivalent Binary Trees
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func flipEquiv(root1 *TreeNode, root2 *TreeNode) bool {
	var dfs func(root1, root2 *TreeNode) bool
	dfs = func(root1, root2 *TreeNode) bool {
		if root1 == root2 || (root1 == nil && root2 == nil) {
			return true
		}
		if root1 == nil || root2 == nil || root1.Val != root2.Val {
			return false
		}
		return (dfs(root1.Left, root2.Left) && dfs(root1.Right, root2.Right)) || (dfs(root1.Left, root2.Right) && dfs(root1.Right, root2.Left))
	}
	return dfs(root1, root2)
}

# Accepted solution for LeetCode #951: Flip Equivalent Binary 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 flipEquiv(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
        def dfs(root1, root2):
            if root1 == root2 or (root1 is None and root2 is None):
                return True
            if root1 is None or root2 is None or root1.val != root2.val:
                return False
            return (dfs(root1.left, root2.left) and dfs(root1.right, root2.right)) or (
                dfs(root1.left, root2.right) and dfs(root1.right, root2.left)
            )

        return dfs(root1, root2)

// Accepted solution for LeetCode #951: Flip Equivalent Binary Trees
struct Solution;
use rustgym_util::*;

trait FlipEq {
    fn flip_eq(root1: &TreeLink, root2: &TreeLink) -> bool;
}

impl FlipEq for TreeLink {
    fn flip_eq(root1: &TreeLink, root2: &TreeLink) -> bool {
        if let (Some(node1), Some(node2)) = (root1, root2) {
            let val1 = node1.borrow().val;
            let val2 = node2.borrow().val;
            let left1 = &node1.borrow().left;
            let right1 = &node1.borrow().right;
            let left2 = &node2.borrow().left;
            let right2 = &node2.borrow().right;
            val1 == val2
                && (Self::flip_eq(left1, left2) && Self::flip_eq(right1, right2)
                    || Self::flip_eq(left1, right2) && Self::flip_eq(right1, left2))
        } else {
            root1 == root2
        }
    }
}

impl Solution {
    fn flip_equiv(root1: TreeLink, root2: TreeLink) -> bool {
        TreeLink::flip_eq(&root1, &root2)
    }
}

#[test]
fn test() {
    let root1 = tree!(
        1,
        tree!(2, tree!(4), tree!(5, tree!(7), tree!(8))),
        tree!(3, tree!(6), None)
    );
    let root2 = tree!(
        1,
        tree!(3, None, tree!(6)),
        tree!(2, tree!(4), tree!(5, tree!(8), tree!(7)))
    );
    let res = true;
    assert_eq!(Solution::flip_equiv(root1, root2), res);
    let root1 = tree!(0, tree!(3), tree!(1, None, tree!(2)));
    let root2 = tree!(0, tree!(3, tree!(2), None), tree!(1));
    let res = false;
    assert_eq!(Solution::flip_equiv(root1, root2), res);
}

// Accepted solution for LeetCode #951: Flip Equivalent Binary Trees
function flipEquiv(root1: TreeNode | null, root2: TreeNode | null): boolean {
    if (root1 === root2) return true;
    if (!root1 || !root2 || root1?.val !== root2?.val) return false;

    const { left: l1, right: r1 } = root1!;
    const { left: l2, right: r2 } = root2!;

    return (flipEquiv(l1, l2) && flipEquiv(r1, r2)) || (flipEquiv(l1, r2) && flipEquiv(r1, l2));
}

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(h)

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.