# LC 662. Maximum Width of Binary Tree

### [Problem link](https://leetcode.com/problems/maximum-width-of-binary-tree/)

###### tags: `leedcode` `python` `medium` `BFS`

Given the <code>root</code> of a binary tree, return the **maximum width** of the given tree.

The **maximum width** of a tree is the maximum **width** among all levels.

The **width** of one level is defined as the length between the end-nodes (the leftmost and rightmost non-null nodes), where the null nodes between the end-nodes that would be present in a complete binary tree extending down to that level are also counted into the length calculation.

It is **guaranteed** that the answer will in the range of a **32-bit** signed integer.

**Example 1:**
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/03/width1-tree.jpg" style="width: 359px; height: 302px;" />

Input: root = [1,3,2,5,3,null,9]
Output: 4
Explanation: The maximum width exists in the third level with length 4 (5,3,null,9).

**Example 2:**
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/14/maximum-width-of-binary-tree-v3.jpg" style="width: 442px; height: 422px;" />

Input: root = [1,3,2,5,null,null,9,6,null,7]
Output: 7
Explanation: The maximum width exists in the fourth level with length 7 (6,null,null,null,null,null,7).

**Example 3:**
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/03/width3-tree.jpg" style="width: 289px; height: 299px;" />

Input: root = [1,3,2,5]
Output: 2
Explanation: The maximum width exists in the second level with length 2 (3,2).

**Constraints:**

- The number of nodes in the tree is in the range <code>[1, 3000]</code>.
- <code>-100 <= Node.val <= 100</code>

## Solution 1 - BFS

# 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 widthOfBinaryTree(self, root: Optional[TreeNode]) -> int:
        q = deque([(root, 1)])
        res = 1

        while q:
            size_q = len(q)
            level_max_width = q[-1][1] - q[0][1] + 1
            res = max(res, level_max_width)
            
            for i in range(size_q):
                node, number = q.popleft()
                if node.left:
                    q.append((node.left, number * 2))
                if node.right:
                    q.append((node.right, number * 2 + 1))

        return res

>### Complexity
>n = The number of nodes in the tree.
>w = Max width of the tree.
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(w) |

## Note
標準的BFS題目, 用iteration就輕鬆解決.