import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

/**

 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.

 */

public class Average_of_Levels_in_Binary_Tree {

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public List<Double> averageOfLevels(TreeNode root) {

            List<Double> result = new ArrayList<>();

            // bfs
            if (root == null) {
                return result;
            }

            Queue<TreeNode> q = new LinkedList<>();
            q.offer(root);
            int currentLevelCount = 1;
            int nextLevelCount = 0;

            long currentLevelSum = 0;
            int count = 0;

            boolean isLevelLeftMost = false;

            while (!q.isEmpty()) {
                TreeNode current = q.poll();
                currentLevelCount--;
                count++;

                /*

                overflow:
                Input:
                        [2147483647,2147483647,2147483647]
                Output:
                           [2147483647.0,-1.0]



                short: 32767 or 0x7fff
                int: 2147483647 or 0x7fffffff
                size_t: 18446744073709551615 or 0xffffffffffffffff
                streamsize: 9223372036854775807 or 0x7fffffffffffffff
                float: 3.40282e+38 or 0x1.fffffep+127
                double: 1.79769e+308 or 0x1.fffffffffffffp+1023

                */


                // @note:@memorize: overflow
                // change sum from int to long
                currentLevelSum += current.val;

                if (current.left != null) {
                    q.offer(current.left);
                    nextLevelCount++;
                }
                if (current.right != null) {
                    q.offer(current.right);
                    nextLevelCount++;
                }

                if (currentLevelCount == 0) {
                    currentLevelCount = nextLevelCount;
                    nextLevelCount = 0;

                    // calculate avg
                    result.add(currentLevelSum * 1.0 / count);

                    count = 0;
                    currentLevelSum = 0;
                }
            }

            return result;
        }
    }
}

############

/**
 * 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;
    }
}

// OJ: https://leetcode.com/problems/average-of-levels-in-binary-tree/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<double> averageOfLevels(TreeNode* root) {
        if (!root) return {};
        vector<double> ans;
        queue<TreeNode*> q;
        q.push(root);
        while (q.size()) {
            double cnt = q.size(), sum = 0;
            for (int i = 0; i < cnt; ++i) {
                root = q.front();
                q.pop();
                sum += root->val;
                if (root->left) q.push(root->left);
                if (root->right) q.push(root->right);
            }
            ans.push_back(sum / cnt);
        }
        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.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
from collections import deque


class Solution(object):
  def averageOfLevels(self, root):
    """
    :type root: TreeNode
    :rtype: List[float]
    """
    ans = []
    queue = deque([root])
    while queue:
      s = 0
      n = len(queue)
      for _ in range(n):
        top = queue.popleft()
        s += top.val
        if top.left:
          queue.append(top.left)
        if top.right:
          queue.append(top.right)
      ans.append(float(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
    }
}

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

############

/**
 * 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;
    }
}

// OJ: https://leetcode.com/problems/average-of-levels-in-binary-tree/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<double> averageOfLevels(TreeNode* root) {
        if (!root) return {};
        vector<double> ans;
        queue<TreeNode*> q;
        q.push(root);
        while (q.size()) {
            double cnt = q.size(), sum = 0;
            for (int i = 0; i < cnt; ++i) {
                root = q.front();
                q.pop();
                sum += root->val;
                if (root->left) q.push(root->left);
                if (root->right) q.push(root->right);
            }
            ans.push_back(sum / cnt);
        }
        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.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
from collections import deque


class Solution(object):
  def averageOfLevels(self, root):
    """
    :type root: TreeNode
    :rtype: List[float]
    """
    ans = []
    queue = deque([root])
    while queue:
      s = 0
      n = len(queue)
      for _ in range(n):
        top = queue.popleft()
        s += top.val
        if top.left:
          queue.append(top.left)
        if top.right:
          queue.append(top.right)
      ans.append(float(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
    }
}

637 - Average of Levels in Binary Tree

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