637. 二叉树的层平均值

English Version

题目描述

给定一个非空二叉树的根节点 root , 以数组的形式返回每一层节点的平均值。与实际答案相差 10-5 以内的答案可以被接受。

 

示例 1：

输入：root = [3,9,20,null,null,15,7]
输出：[3.00000,14.50000,11.00000]
解释：第 0 层的平均值为 3,第 1 层的平均值为 14.5,第 2 层的平均值为 11 。
因此返回 [3, 14.5, 11] 。

示例 2:

输入：root = [3,9,20,15,7]
输出：[3.00000,14.50000,11.00000]

 

提示：

• 树中节点数量在 [1, 104] 范围内
• -231 <= Node.val <= 231 - 1

解法

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def averageOfLevels(self, root: TreeNode) -> List[float]:
        res = []
        q = deque([root])
        while q:
            n = len(q)
            s = 0
            for _ in range(n):
                node = q.popleft()
                s += node.val
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            res.append(s / n)
        return res

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> res = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        while (!q.isEmpty()) {
            double s = 0, n = q.size();
            for (int i = 0; i < n; ++i) {
                TreeNode node = q.poll();
                s += node.val;
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
            res.add(s / n);
        }
        return res;
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var averageOfLevels = function (root) {
    let res = [];
    let queue = [root];
    while (queue.length > 0) {
        n = queue.length;
        let sum = 0;
        for (let i = 0; i < n; i++) {
            let node = queue.shift();
            sum += node.val;
            if (node.left) {
                queue.push(node.left);
            }
            if (node.right) {
                queue.push(node.right);
            }
        }
        res.push(sum / n);
    }
    return res;
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func averageOfLevels(root *TreeNode) []float64 {
	q := []*TreeNode{root}
	var ans []float64
	for len(q) > 0 {
		n := len(q)
		var sum int
		for i := 0; i < n; i++ {
			node := q[0]
			q = q[1:]
			sum += node.Val
			if node.Left != nil {
				q = append(q, node.Left)
			}
			if node.Right != nil {
				q = append(q, node.Right)
			}
		}
		ans = append(ans, float64(sum)/float64(n))
	}
	return ans
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<double> averageOfLevels(TreeNode* root) {
        queue<TreeNode*> q({root});
        vector<double> ans;
        while (!q.empty()) {
            int n = q.size();
            long long sum = 0;
            for (int i = 0; i < n; ++i) {
                TreeNode* node = q.front();
                q.pop();
                sum += node->val;
                if (node->left != nullptr) q.push(node->left);
                if (node->right != nullptr) q.push(node->right);
            }
            ans.emplace_back(sum * 1.0 / n);
        }
        return ans;
    }
};

Rust

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;
impl Solution {
    pub fn average_of_levels(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<f64> {
        if root.is_none() {
            return Vec::new();
        }

        let mut q = VecDeque::new();
        q.push_back(Rc::clone(&root.unwrap()));
        let mut ans = Vec::new();
        while !q.is_empty() {
            let n = q.len();
            let mut sum = 0.0;
            for _ in 0..n {
                let node = q.pop_front().unwrap();
                sum += node.borrow().val as f64;
                if node.borrow().left.is_some() {
                    q.push_back(Rc::clone(node.borrow().left.as_ref().unwrap()));
                }
                if node.borrow().right.is_some() {
                    q.push_back(Rc::clone(node.borrow().right.as_ref().unwrap()));
                }
            }
            ans.push(sum / n as f64);
        }
        ans
    }
}

...