leetcode515- Find Largest Value in Each Tree Row- medium

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]

算法：

1.BFS。层级遍历，每层打擂台。

2.DFS。记录从上到下的深度，每层参数里传下去+1即可。（从下到上的深度是通过回传int返回值来递归实现的）。每次根据当前深度d跟List<Integer>里对应位置的值打擂台。注意List的set用法。list.set(d, Math.max(list.get(d), n.val));

实现：

1.BFS

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

2.DFS

public class Solution {
    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);
    }
}