LeetCode 517 - Find Largest Value in Each Tree Row

https://leetcode.com/problems/find-largest-value-in-each-tree-row/

You need to find the largest value in each row of a binary tree.

Example:

Input: 

          1
         / \
        3   2
       / \   \  
      5   3   9 

Output: [1, 3, 9]

X. BFS
https://discuss.leetcode.com/topic/78945/java-bfs
https://discuss.leetcode.com/topic/78955/verbose-java-solution-binary-tree-level-order-traversal-again

    public int[] findValueMostElement(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        List<Integer> res = new ArrayList<Integer>();
        queue.add(root);
        int queueSize = root == null ? 0 : 1;
        while (queueSize > 0) {
            int largestElement = Integer.MIN_VALUE;
            for (int i=0;i<queueSize;i++) {
                TreeNode cur = queue.poll();
                largestElement = Math.max(cur.val, largestElement);
                if (cur.left != null) queue.add(cur.left);
                if (cur.right != null) queue.add(cur.right);
            }
            res.add(largestElement);
            queueSize = queue.size();
        }
        int[] resArray = new int[res.size()];
        for (int i=0;i<res.size();i++) resArray[i] = res.get(i);
        return resArray;
    }

X. DFS
https://discuss.leetcode.com/topic/79178/9ms-java-dfs-solution

Just a simple pre-order traverse idea. Use depth to expand result list size and put the max value in the appropriate position.

    public List<Integer> largestValues(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        helper(root, res, 0);
        return res;
    }
    private void helper(TreeNode root, List<Integer> res, int d){
        if(root == null){
            return;
        }
       //expand list size
        if(d == res.size()){
            res.add(root.val);
        }
        else{
        //or set value
            res.set(d, Math.max(res.get(d), root.val));
        }
        helper(root.left, res, d+1);
        helper(root.right, res, d+1);
    }