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637. Average of Levels in Binary Tree

Description

Given the root of a binary tree, return the average value of the nodes on each level in the form of an array. Answers within 10-5 of the actual answer will be accepted.

 

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [3.00000,14.50000,11.00000]
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].

Example 2:

Input: root = [3,9,20,15,7]
Output: [3.00000,14.50000,11.00000]

 

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • -231 <= Node.val <= 231 - 1

Solutions

  • /**
 * 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> ans = new ArrayList<>();
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            int n = q.size();
            long s = 0;
            for (int i = 0; i < n; ++i) {
                root = q.pollFirst();
                s += root.val;
                if (root.left != null) {
                    q.offer(root.left);
                }
                if (root.right != null) {
                    q.offer(root.right);
                }
            }
            ans.add(s * 1.0 / n);
        }
        return ans;
    }
}

  • /**
 * 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 s = 0;
            for (int i = 0; i < n; ++i) {
                root = q.front();
                q.pop();
                s += root->val;
                if (root->left) q.push(root->left);
                if (root->right) q.push(root->right);
            }
            ans.push_back(s * 1.0 / n);
        }
        return ans;
    }
};

  • # 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: Optional[TreeNode]) -> List[float]:
        q = deque([root])
        ans = []
        while q:
            s, n = 0, len(q)
            for _ in range(n):
                root = q.popleft()
                s += root.val
                if root.left:
                    q.append(root.left)
                if root.right:
                    q.append(root.right)
            ans.append(s / n)
        return ans


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

  • /**
 * 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 q = [root];
    let ans = [];
    while (q.length) {
        const n = q.length;
        let s = 0;
        for (let i = 0; i < n; ++i) {
            root = q.shift();
            s += root.val;
            if (root.left) {
                q.push(root.left);
            }
            if (root.right) {
                q.push(root.right);
            }
        }
        ans.push(s / n);
    }
    return ans;
};


  • // 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
    }
}

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