• [LeetCode] Average of Levels in Binary Tree 二叉树的层平均值


    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.

    这道题让我们求一个二叉树每层的平均值,那么一看就是要进行层序遍历了,直接上queue啊,如果熟悉层序遍历的方法,那么这题就没有什么难度了,直接将每层的值累计加起来,除以该层的结点个数,存入结果res中即可,参见代码如下:

    解法一:

    class Solution {
    public:
        vector<double> averageOfLevels(TreeNode* root) {
            if (!root) return {};
            vector<double> res;
            queue<TreeNode*> q{{root}};
            while (!q.empty()) {
                int n = q.size();
                double sum = 0;
                for (int i = 0; i < n; ++i) {
                    TreeNode *t = q.front(); q.pop();
                    sum += t->val;
                    if (t->left) q.push(t->left);
                    if (t->right) q.push(t->right);
                }
                res.push_back(sum / n);
            }
            return res;
        }
    };

    下面这种方法虽然是利用的递归形式的先序遍历,但是其根据判断当前层数level跟结果res中已经初始化的层数之间的关系对比,能把当前结点值累计到正确的位置,而且该层的结点数也自增1,这样我们分别求了两个数组,一个数组保存了每行的所有结点值,另一个保存了每行结点的个数,这样对应位相除就是我们要求的结果了,参见代码如下:

    解法二:

    class Solution {
    public:
        vector<double> averageOfLevels(TreeNode* root) {
            vector<double> res, cnt;
            helper(root, 0, cnt, res);
            for (int i = 0; i < res.size(); ++i) {
                res[i] /= cnt[i];
            }
            return res;
        }
        void helper(TreeNode* node, int level, vector<double>& cnt, vector<double>& res) {
            if (!node) return;
            if (res.size() <= level) {
                res.push_back(0);
                cnt.push_back(0);
            }
            res[level] += node->val;
            ++cnt[level];
            helper(node->left, level + 1, cnt, res);
            helper(node->right, level + 1, cnt, res);
        }
    };

    类似题目:

    Binary Tree Level Order Traversal II

    Binary Tree Level Order Traversal

    参考资料:

    https://discuss.leetcode.com/topic/95567/java-solution-using-dfs-with-full-comments

     

    LeetCode All in One 题目讲解汇总(持续更新中...) 

  • 相关阅读:
    覆盖式发布与非覆盖式发布
    GIT
    Web Service返回符合Xml Schema规范的Xml文档
    下拉渐显菜单
    计算网页上坐标的距离
    初识交互设计
    良好用户体验-实现过程!
    做 用户调研?
    这个没什么技术含量,实现起来很简单?
    SQL SERVER 登录问题!该用户与可信的Sql Server连接无关联
  • 原文地址:https://www.cnblogs.com/grandyang/p/7259209.html
Copyright © 2020-2023  润新知