637. Average of Levels in Binary Tree

心已赠人 2022-06-02 00:26 111阅读 0赞

[原题链接][Link 1]

主要考察的是对树这个数据结构的遍历。这里有两种方式

**BFS**  
在这种题目上来说，使用BFS进行层次遍历进行代码的书写更舒服

public List<Double> averageOfLevels(TreeNode root) {
      List<Double> list = new LinkedList<>();
      Queue<TreeNode> queue = new LinkedList<>();
      queue.offer(root);
    
      while (!queue.isEmpty()) {
          int count = queue.size();
          double sum = 0;
          for (int i = 0; i < count; i++) {
              TreeNode cur = queue.poll();
              sum += cur.val;
              if (cur.left != null) queue.offer(cur.left);
              if (cur.right != null) queue.offer(cur.right);
          }
          list.add(sum / count);
      }
      return list;
    }

--------------------

**DFS**  
如果使用深度优先遍历则需要借助额外的数据结构

public List<Double> averageOfLevelsByDfs(TreeNode root) {
        List<Node> temp = new ArrayList<>();
    
        helper(root, temp, 0);
        List<Double> result = new LinkedList<>();
        for (int i = 0; i < temp.size(); i++) {
            result.add(temp.get(i).sum / temp.get(i).count);
        }
        return result;
    }
    public void helper(TreeNode root, List<Node> temp, int level) {
        if (root == null) return;
        //因为level是从0开始的， 如果level等于当前的temp.size()。说明进入了新的level，需要加入新的数组元素
        if (level == temp.size()) {
            Node node = new Node((double)root.val, 1);
            temp.add(node);
        } else {
            temp.get(level).sum += root.val;
            temp.get(level).count++;
        }
        helper(root.left, temp, level + 1);
        helper(root.right, temp, level + 1);
    }
    
    class Node { 
        double sum;
        int count;
    
        Node(double sum, int count) {
            this.sum = sum;
            this.count = count;
        }
    }

[Link 1]: https://leetcode.com/problems/average-of-levels-in-binary-tree/description/