## [662\. Maximum Width of Binary Tree](https://leetcode.com/problems/maximum-width-of-binary-tree/)

Given the `root` 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:**


**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:**


**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:**


**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 `[1, 3000]`.
- `-100 <= Node.val <= 100`

/**
 * 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) {}
 * };
 */
 #define ll long long

class Solution {
public:
    int widthOfBinaryTree(TreeNode* root) {
        if(!root) return 0;
        // 使用 pair 的結構 {目前的節點, 節點的 index}
        queue<pair<TreeNode*, ll>> q;
        q.push({root, 0});

        ll maxWidth = 0;
        while(!q.empty()) {
            ll qSize = q.size();
            // 最左邊 node 的 index
            ll left = q.front().second;
            // 最右邊 node 的 index
            ll right = q.back().second;
            
            // 寬度公式：右邊節點的 index - 左邊節點的 index + 1
            maxWidth = max(maxWidth, right - left + 1);

            // 走訪每一層
            for(int i = 0; i < qSize; i++) {
                auto node = q.front(); q.pop();
                // levelIndex = 目前節點的 index - 最左邊節點的 index
                ll levelIndex = node.second - left;
                if(node.first->left) {
                    // 將左邊節點推到 queue
                    q.push({node.first->left, 2 * levelIndex});
                }
                if(node.first->right) {
                    // 將右邊節點推到 queue
                    q.push({node.first->right, 2 * levelIndex + 1});
                }                
            }
        }
        return maxWidth;
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::