515. Find Largest Value in Each Tree Row

Version #1 BFS
Using a temporary max variable to keep track of the maximum value of each level
71.59%

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> largestValues(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) return result;
        Queue<TreeNode> que = new LinkedList<>();
        que.add(root);
        int size, max;
        TreeNode curr = null;
        while (!que.isEmpty()) {
            size = que.size();
            max = Integer.MIN_VALUE;
            for (int i = 0; i < size; i++) {
                curr = que.poll();
                max = Math.max(curr.val, max);
                if (curr.left != null) que.offer(curr.left);
                if (curr.right != null) que.offer(curr.right);
            }
            result.add(max);
        }
        return result;
    }
}


Version #2 DFS

 95.56%
public class Solution {
    public List<Integer> largestValues(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) return result;
        dfs(root, 0, result);
        return result;
    }
 
    private void dfs(TreeNode root, int level, List<Integer> result) {
        //Base case
        if (root == null) return;
        //1st time to touch this level
        if (level == result.size()) {
            result.add(root.val);
        //The level maximum needs to be updated
        } else if (root.val > result.get(level)) {
            result.set(level, root.val);
        }
        //Do recursion no matter its children are null or not
        //Base case takes care of null
        dfs(root.left, level + 1, result);
        dfs(root.right, level + 1, result);
    }
}