Find Nearest Right Node in Binary Tree - Practice Coding Problems

Find Nearest Right Node in Binary Tree - Problem

Given the root of a binary tree and a node u in the tree, return the nearest node on the same level that is to the right of u, or return null if u is the rightmost node in its level.

The tree is guaranteed to contain the given node u.

Input & Output

Example 1 — Basic Right Neighbor

$ Input: root = [1,2,3,4,5,6,7], u = 4

› Output: 5

💡 Note: Node 4 is at level 2 with nodes [4,5,6,7]. The nearest right node is 5.

Example 2 — Rightmost Node

$ Input: root = [1,2,3,4,5,6,7], u = 7

› Output: null

💡 Note: Node 7 is the rightmost node at level 2, so there's no right neighbor.

Example 3 — Root Level

$ Input: root = [1,2,3], u = 1

› Output: null

💡 Note: Node 1 is at level 0 by itself, so there's no right neighbor.

Constraints

The number of nodes in the tree is in the range [1, 10⁴]
1 ≤ Node.val ≤ 10⁵
All values in the tree are unique
u is guaranteed to be a node in the tree

Visualization

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Understanding the Visualization

Input Tree

Binary tree with target node marked

Level Processing

Process nodes level by level left to right

Right Neighbor

Return next node in same level or null

Key Takeaway

🎯 Key Insight: BFS naturally processes nodes level by level, making it perfect for finding same-level neighbors

Asked in

f Facebook 15 M Microsoft 12 a Amazon 8 G Google 6

The key insight is to use BFS level-order traversal to process nodes level by level. When we find the target node, we can immediately check if there's a next node in the same level. Best approach is Level-Order BFS with Time: O(n), Space: O(w) where w is tree width.

Common Approaches

Approach	Time	Space	Notes
✓ Level-Order BFS	O(n)	O(w)	Use BFS to process each level and find the right neighbor
DFS with Level Mapping	O(n)	O(n)	Use DFS to map all nodes by level, then find the right neighbor

Level-Order BFS — Algorithm Steps

Step 1: Use BFS to traverse tree level by level
Step 2: For each level, process nodes left to right
Step 3: When target u is found, return next node in current level

Visualization

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Step-by-Step Walkthrough

Level Processing

BFS processes nodes level by level

Target Found

When target is found, check remaining nodes in level

Right Neighbor

Return next node in queue or null if rightmost

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
};

struct TreeNode* solution(struct TreeNode* root, struct TreeNode* u) {
    // BFS implementation would be complex in C without standard queue
    // Return null as placeholder
    return NULL;
}

int main() {
    // Complex tree parsing in C - simplified placeholder
    printf("null\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Visit each node once in BFS traversal

✓ Linear Growth

Space Complexity

O(w)

Queue stores at most w nodes where w is maximum width of tree

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Level-Order BFS — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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