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  • 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.
     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     int val;
     5  *     TreeNode left;
     6  *     TreeNode right;
     7  *     TreeNode(int x) { val = x; }
     8  * }
     9  */
    10 public class Solution {
    11     public List<Double> averageOfLevels(TreeNode root) {
    12         List<Double> average = new ArrayList<>();
    13         if (root == null) return average;
    14         
    15         Queue<TreeNode> queue = new LinkedList<>();
    16         queue.add(root);
    17         while (queue.peek() != null) {
    18             average.add(0.0);
    19             int n = average.size() - 1;
    20             
    21             int levelCount = queue.size();
    22             int m = levelCount;
    23             while (levelCount-- != 0) {
    24                 TreeNode temp = queue.poll();
    25                 average.set(n, average.get(n) + temp.val);
    26                 
    27                 if (temp.left != null) queue.add(temp.left);
    28                 if (temp.right != null) queue.add(temp.right);
    29             }
    30             average.set(n, average.get(n) / m);
    31         }
    32         return average;
    33     }
    34 }


     

     
     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     int val;
     5  *     TreeNode left;
     6  *     TreeNode right;
     7  *     TreeNode(int x) { val = x; }
     8  * }
     9  */
    10 public class Solution {
    11     public List<Double> averageOfLevels(TreeNode root) {
    12         List<Double> sum = new ArrayList<Double>();
    13         List<Integer> count = new ArrayList<Integer>();
    14         
    15         levelTraversal(root, sum, count, 0);
    16         countAverage(sum, count);
    17         
    18         return sum;
    19     }
    20     
    21     private void levelTraversal(TreeNode root, List<Double> sum, List<Integer> count, int depth) {
    22         if (root == null) return;
    23         if (depth == sum.size()) {
    24             sum.add(0.0);
    25             count.add(0);
    26         }
    27         
    28         sum.set(depth, sum.get(depth) + root.val);
    29         count.set(depth, count.get(depth) + 1);
    30         
    31         levelTraversal(root.left, sum, count, depth + 1);
    32         levelTraversal(root.right, sum, count, depth + 1);
    33     }
    34     
    35     private void countAverage(List<Double> sum, List<Integer> count) {
    36         int n = sum.size();
    37         for (int i = 0; i < n; i++) {
    38             sum.set(i, sum.get(i) * 1.0 / count.get(i));
    39         }
    40     }
    41 }
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  • 原文地址:https://www.cnblogs.com/amazingzoe/p/7145186.html
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