LeetCode 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.

Answer:

/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var averageOfLevels = function(root) {
    if(root == null) {
        return 0;
    }
    
    var queue = [];var arr = [];
    
    queue.push(root);
    
    while(queue.length != 0) {        
        var sum = 0; var count = 0;
        var temp = [];
        while(queue.length != 0) {
            var firstNode = queue.splice(0,1)[0];
            sum+= firstNode.val;                     
            if(firstNode.left)
                temp.push(firstNode.left);
            if(firstNode.right) {
                temp.push(firstNode.right);
            }
            count++;        
        }
        queue = temp;
        arr.push(sum/ count);
    }
    
    console.log(arr);
    return arr;
    
};

function TreeNode(val) {
    this.val = val;
    this.left = this.right = null;
}

var rootNode = new TreeNode(3);
var lrNode = new TreeNode(9);
var rrNode = new TreeNode(20);
var lrrNode = new TreeNode(15);
var rrrNode = new TreeNode(7);
rrNode.right = rrrNode;
rrNode.left = lrrNode;
rootNode.left = lrNode;
rootNode.right = rrNode;

averageOfLevels(rootNode);

 

这里应用BFS的解法。网上还有一种DFS的解法不过还没有研究，就不写了。