515. 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]
题目含义：从上到下找到每行的最大值
方法一：DFS
 参数d表示层数的索引，第一行对应的d为0

 1     private void largestValue(TreeNode root, List<Integer> res, int d){
 2         if(root == null){
 3             return;
 4         }
 5         //expand list size
 6         if(d == res.size()){
 7             res.add(root.val); //在最下面添加新的行
 8         }
 9         else{
10             //or set value
11             res.set(d, Math.max(res.get(d), root.val)); //当前节点值处于第(d+1)层，如果当前值大于res中保存的值，则替换成最大值
12         }
13         largestValue(root.left, res, d+1);
14         largestValue(root.right, res, d+1);
15     }
16     
17     
18     public List<Integer> largestValues(TreeNode root) {
19         List<Integer> res = new ArrayList<Integer>();
20         largestValue(root, res, 0);
21         return res;     
22     }

方法二：

 1     public List<Integer> largestValues(TreeNode root) {
 2         Queue<TreeNode> queue = new LinkedList<TreeNode>();
 3         List<Integer> res = new ArrayList<Integer>();
 4         if (root == null) return res;
 5         queue.add(root);
 6         while (!queue.isEmpty()) {
 7             int largestElement = Integer.MIN_VALUE;
 8             int queueSize = queue.size();
 9             for (int i=0;i<queueSize;i++) {
10                 TreeNode cur = queue.poll();
11                 largestElement = Math.max(cur.val, largestElement);
12                 if (cur.left != null) queue.add(cur.left);
13                 if (cur.right != null) queue.add(cur.right);
14             }
15             res.add(largestElement);
16         }
17         return res;
18     }