637. Average of Levels in Binary Tree

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:

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

class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> result = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();

        if(root == null) return result;
        q.add(root);
        while(!q.isEmpty()) {
            int n = q.size();
            double sum = 0.0;
            for(int i = 0; i < n; i++) {
                TreeNode node = q.poll();
                sum += node.val;
                if(node.left != null) q.offer(node.left);
                if(node.right != null) q.offer(node.right);
            }
            result.add(sum / n);
        }
        return result;
    }
}

BFS, 注意Integer.MAX_VALUE, 所以要用double在里面

class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        List<List<Integer>> res = new ArrayList();
        dfs(res, 0, root);
        List<Double> result = new ArrayList();
        for(List<Integer> list: res){
            double cur = 0;
            for(int i: list){
                cur+=i;
            }
            result.add(cur / (double) list.size());
        }
        return result;
    }
    public void dfs(List<List<Integer>> res, int h, TreeNode root){
        if(root == null) return;
        if(res.size() == h) res.add(new ArrayList());
        res.get(h).add(root.val);
        dfs(res, h+1, root.left);
        dfs(res, h+1, root.right);
    }
}

DFS，和level order traverse一样

Just like all other level order traverse