【LeetCode】637. Average of Levels in Binary Tree

最新推荐文章于 2021-08-07 10:51:15 发布

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最新推荐文章于 2021-08-07 10:51:15 发布

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本文介绍了一种计算非空二叉树各层级节点平均值的方法，通过两种不同的算法实现：深度优先搜索（DFS）和广度优先搜索（BFS）。深度优先搜索通过递归遍历树的每个节点来收集数据，而广度优先搜索则按层级遍历。最终输出各层级的平均值。

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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.

//深度优先算法
var average = function(t,  i, sum, count) {
        if (t == null)
            return;
        if (i < sum.length) {
            sum.splice(i,1,sum[i] + t.val);
            count.splice(i,1,count[i] + 1);
        } else {
            sum.push(1.0 * t.val);
            count.push(1);
        }
        average(t.left, i + 1, sum, count);
        average(t.right, i + 1, sum, count);
    }
var averageOfLevels = function(root) {
    var count = new Array();
    var res = new Array();
    average(root, 0, res, count);
    for (var i = 0; i < res.length; i++)
        res.splice(i,1, res[i] / count[i]);
    return res;
}

//广度优先算法
var averageOfLevels = function(root) {
    var res=new Array();
    var queue=new Array();
    queue.push(root);
    while(queue.length>0){
        var sum=0,count=0;
        var temp=new Array();
        while(queue.length>0){
            var n=queue.shift();
            sum+= n.val;
            count++;
            if(n.left!=null){
                temp.push(n.left);
            }
             if(n.right!=null){
                temp.push(n.right);
            }
        }
        queue=temp;
        res.push(sum/count*1.0);
    }
    return res;
};