637. Average of Levels in Binary Tree.md

637. Average of Levels in Binary Tree

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leetcode-cn Daily Challenge on September 12th, 2020.

leetcode Daily Challenge on March 5th, 2021.

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Difficulty : Tree

Related Topics : Easy

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Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.

Example 1:

Input:
    3
   / \
  9  20
    /  \
   15   7
Output: [3, 14.5, 11]
Explanation:
The average value of nodes on level 0 is 3,  on level 1 is 14.5, and on level 2 is 11. Hence return [3, 14.5, 11].

Note:

• The range of node's value is in the range of 32-bit signed integer.

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Solution

• mine
  • Java
    • Runtime: 2 ms, faster than 82.50%, Memory Usage: 41.9 MB, less than 36.35% of Java online submissions

      // O(N)time
      // O(D)space
      public List<Double> averageOfLevels(TreeNode root) {
          Map<Integer,Double> map = new HashMap<>();
          Map<Integer,Integer> count = new HashMap<>();
          dfs(root, map, count, 0);
          List<Double> res = new LinkedList<>();
          int t = 0;
          while(map.containsKey(t)){
              double v = map.get(t);
              int c = count.get(t);
              res.add(v / c);
              t++;
          }
          return res;
      }
      
      void dfs(TreeNode node, Map<Integer,Double> map, Map<Integer,Integer> count,int deep){
          if(node == null) return;
          map.put(deep, map.getOrDefault(deep, 0D) + node.val);
          count.put(deep, count.getOrDefault(deep, 0) + 1);
          dfs(node.left, map, count, deep + 1);
          dfs(node.right, map, count, deep + 1);
      }
      

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