# 34. 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 are also counted into the length calculation.

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

**Example 1:**![](https://assets.leetcode.com/uploads/2021/05/03/width1-tree.jpg)

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

**Example 2:**![](https://assets.leetcode.com/uploads/2021/05/03/width2-tree.jpg)

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

**Example 3:**![](https://assets.leetcode.com/uploads/2021/05/03/width3-tree.jpg)

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

**Example 4:**![](https://assets.leetcode.com/uploads/2021/05/03/width4-tree.jpg)

```
Input: root = [1,3,2,5,null,null,9,6,null,null,7]
Output: 8
Explanation: The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).
```

## Solution: 

**Approach:**\
**Left Child at pos = 2\*i +1** \
**Right Child at pos = 2\*i + 2**\
**Finding the distance between the 1st left and last right of a level** 

```cpp
class Solution
{
public:
    int widthOfBinaryTree(TreeNode *root)
    {

        if (root == NULL)
        {
            return 0;
        }

        queue<TreeNode *> q;
        queue<int> idx;

        int maxWidth = 0;
        q.push(root);
        idx.push(0);

        while (!q.empty())
        {

            int s = q.size();
            int i = 0;
            int start = 0;
            int end = 0;
            int minVal = idx.front();

            while (i < s)
            {
                int pos = idx.front() - minVal;
                idx.pop();

                if (i == 0)
                {
                    start = pos - minVal;
                }

                if (i == s - 1)
                {
                    end = pos - minVal;
                }

                TreeNode *t = q.front();
                q.pop();

                if (t->left)
                {
                    q.push(t->left);
                    idx.push(2 * pos + 1);
                }

                if (t->right)
                {
                    q.push(t->right);
                    idx.push(2 * pos + 2);
                }

                i++;
            }

            maxWidth = max(maxWidth, (end - start) + 1);
        }

        return maxWidth;
    }
};
```


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