LeetCode 1602: Find Nearest Right Node in Binary Tree Solution

Master LeetCode problem 1602 (Find Nearest Right Node in Binary Tree), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1602. Find Nearest Right Node in Binary Tree

Medium

Tree Breadth-First Search Binary Tree

Problem Explanation

Explanation

Given a binary tree and a node u, we need to find the nearest node on the right side of u. We can perform a level order traversal of the tree and look for the first occurrence of u in each level. If u is found in a level, then the nearest right node will be the next node encountered in that level.

Algorithm:

Perform a level order traversal of the binary tree.
For each level, check if the current node is equal to the given node u.
If u is found in the level, return the next node encountered in that level.
If u is the last node in its level, return null.

Time Complexity: O(N) where N is the number of nodes in the tree. Space Complexity: O(N) for the queue used in level order traversal.

Solution Code

class Solution {
    public TreeNode findNearestRightNode(TreeNode root, TreeNode u) {
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            boolean found = false;
            
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                
                if (found) return node;
                if (node == u) found = true;
                
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
        }
        
        return null;
    }
}

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