LeetCode #2641 — MEDIUM

Cousins in Binary Tree II

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

The Problem

Problem Statement

Given the root of a binary tree, replace the value of each node in the tree with the sum of all its cousins' values.

Two nodes of a binary tree are cousins if they have the same depth with different parents.

Return the root of the modified tree.

Note that the depth of a node is the number of edges in the path from the root node to it.

Example 1:

Input: root = [5,4,9,1,10,null,7]
Output: [0,0,0,7,7,null,11]
Explanation: The diagram above shows the initial binary tree and the binary tree after changing the value of each node.
- Node with value 5 does not have any cousins so its sum is 0.
- Node with value 4 does not have any cousins so its sum is 0.
- Node with value 9 does not have any cousins so its sum is 0.
- Node with value 1 has a cousin with value 7 so its sum is 7.
- Node with value 10 has a cousin with value 7 so its sum is 7.
- Node with value 7 has cousins with values 1 and 10 so its sum is 11.

Example 2:

Input: root = [3,1,2]
Output: [0,0,0]
Explanation: The diagram above shows the initial binary tree and the binary tree after changing the value of each node.
- Node with value 3 does not have any cousins so its sum is 0.
- Node with value 1 does not have any cousins so its sum is 0.
- Node with value 2 does not have any cousins so its sum is 0.

Constraints:

The number of nodes in the tree is in the range [1, 10⁵].
1 <= Node.val <= 10⁴

Step 01

Brute Force Baseline

Problem summary: Given the root of a binary tree, replace the value of each node in the tree with the sum of all its cousins' values. Two nodes of a binary tree are cousins if they have the same depth with different parents. Return the root of the modified tree. Note that the depth of a node is the number of edges in the path from the root node to it.

Baseline thinking

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

Pattern signal: Hash Map · Tree

Example 1

[5,4,9,1,10,null,7]

Example 2

[3,1,2]

Core Insight

What unlocks the optimal approach

Use DFS two times.
For the first time, find the sum of values of all the levels of the binary tree.
For the second time, update the value of the node with the sum of the values of the current level - sibling node’s values.

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 #2641: Cousins in Binary Tree II
/**
 * 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 List<Integer> s = new ArrayList<>();

    public TreeNode replaceValueInTree(TreeNode root) {
        dfs1(root, 0);
        root.val = 0;
        dfs2(root, 0);
        return root;
    }

    private void dfs1(TreeNode root, int depth) {
        if (root == null) {
            return;
        }
        if (s.size() <= depth) {
            s.add(0);
        }
        s.set(depth, s.get(depth) + root.val);
        dfs1(root.left, depth + 1);
        dfs1(root.right, depth + 1);
    }

    private void dfs2(TreeNode root, int depth) {
        int l = root.left == null ? 0 : root.left.val;
        int r = root.right == null ? 0 : root.right.val;
        int sub = l + r;
        ++depth;
        if (root.left != null) {
            root.left.val = s.get(depth) - sub;
            dfs2(root.left, depth);
        }
        if (root.right != null) {
            root.right.val = s.get(depth) - sub;
            dfs2(root.right, depth);
        }
    }
}

// Accepted solution for LeetCode #2641: Cousins in Binary Tree II
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func replaceValueInTree(root *TreeNode) *TreeNode {
	s := []int{}
	var dfs1 func(*TreeNode, int)
	dfs1 = func(root *TreeNode, depth int) {
		if root == nil {
			return
		}
		if len(s) <= depth {
			s = append(s, 0)
		}
		s[depth] += root.Val
		dfs1(root.Left, depth+1)
		dfs1(root.Right, depth+1)
	}
	var dfs2 func(*TreeNode, int)
	dfs2 = func(root *TreeNode, depth int) {
		l, r := 0, 0
		if root.Left != nil {
			l = root.Left.Val
		}
		if root.Right != nil {
			r = root.Right.Val
		}
		sub := l + r
		depth++
		if root.Left != nil {
			root.Left.Val = s[depth] - sub
			dfs2(root.Left, depth)
		}
		if root.Right != nil {
			root.Right.Val = s[depth] - sub
			dfs2(root.Right, depth)
		}
	}
	dfs1(root, 0)
	root.Val = 0
	dfs2(root, 0)
	return root
}

# Accepted solution for LeetCode #2641: Cousins in Binary Tree II
# 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 replaceValueInTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        def dfs1(root: Optional[TreeNode], depth: int):
            if root is None:
                return
            if len(s) <= depth:
                s.append(0)
            s[depth] += root.val
            dfs1(root.left, depth + 1)
            dfs1(root.right, depth + 1)

        def dfs2(root: Optional[TreeNode], depth: int):
            sub = (root.left.val if root.left else 0) + (
                root.right.val if root.right else 0
            )
            depth += 1
            if root.left:
                root.left.val = s[depth] - sub
                dfs2(root.left, depth)
            if root.right:
                root.right.val = s[depth] - sub
                dfs2(root.right, depth)

        s = []
        dfs1(root, 0)
        root.val = 0
        dfs2(root, 0)
        return root

// Accepted solution for LeetCode #2641: Cousins in Binary Tree II
/**
 * [2641] Cousins in Binary Tree II
 */
pub struct Solution {}

use crate::util::tree::{to_tree, TreeNode};

// submission codes start here

// 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 replace_value_in_tree(
        root: Option<Rc<RefCell<TreeNode>>>,
    ) -> Option<Rc<RefCell<TreeNode>>> {
        let root = if let Some(r) = root {
            r
        } else {
            return None;
        };

        let mut queue = Vec::new();
        queue.push(Rc::clone(&root));

        while !queue.is_empty() {
            let mut next_queue = Vec::new();

            let mut sum = 0;
            for node in &queue {
                if let Some(left) = &node.borrow().left {
                    next_queue.push(Rc::clone(left));
                    sum += left.borrow().val;
                }

                if let Some(right) = &node.borrow().right {
                    next_queue.push(Rc::clone(right));
                    sum += right.borrow().val;
                }
            }

            for node in &queue {
                let mut children = 0;
                if let Some(left) = &node.borrow().left {
                    children += left.borrow().val;
                }

                if let Some(right) = &node.borrow().right {
                    children += right.borrow().val;
                }

                if let Some(left) = &node.borrow().left {
                    left.borrow_mut().val = sum - children;
                }

                if let Some(right) = &node.borrow().right {
                    right.borrow_mut().val = sum - children;
                }
            }

            queue = next_queue;
        }

        root.borrow_mut().val = 0;
        Some(root)
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2641() {
        assert_eq!(
            Solution::replace_value_in_tree(to_tree(vec![
                Some(5),
                Some(4),
                Some(9),
                Some(1),
                Some(10),
                None,
                Some(7)
            ])),
            to_tree(vec![
                Some(0),
                Some(0),
                Some(0),
                Some(7),
                Some(7),
                None,
                Some(11)
            ])
        );
    }
}

// Accepted solution for LeetCode #2641: Cousins in Binary Tree II
/**
 * 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 replaceValueInTree(root: TreeNode | null): TreeNode | null {
    const s: number[] = [];
    const dfs1 = (root: TreeNode | null, depth: number) => {
        if (!root) {
            return;
        }
        if (s.length <= depth) {
            s.push(0);
        }
        s[depth] += root.val;
        dfs1(root.left, depth + 1);
        dfs1(root.right, depth + 1);
    };
    const dfs2 = (root: TreeNode | null, depth: number) => {
        const sub = (root.left?.val || 0) + (root.right?.val || 0);
        ++depth;
        if (root.left) {
            root.left.val = s[depth] - sub;
            dfs2(root.left, depth);
        }
        if (root.right) {
            root.right.val = s[depth] - sub;
            dfs2(root.right, depth);
        }
    };
    dfs1(root, 0);
    root.val = 0;
    dfs2(root, 0);
    return root;
}

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.