515. Find Largest Value in Each Tree Row (BFS)

time: O(n)

space: O(D) to keep the queues, where D is a tree diameter. Let's use the last level to estimate the queue size. This level could contain up to N/2N/2 tree nodes in the case of complete binary tree.

/**
 * 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 List<Integer> largestValues(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) return res;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            int max = Integer.MIN_VALUE;
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                max = Math.max(max, node.val);
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
            res.add(max);
        }
        return res;
    }
}