## 662. Maximum Width of Binary Tree

Given a binary tree, write a function to get the maximum width of the given tree. The width of a tree is the maximum width among all levels. The binary tree has the same structure as a full binary tree, but some nodes are null.

The width of one level is defined as the length between the end-nodes (the leftmost and right most non-null nodes in the level, where the `null` nodes between the end-nodes are also counted into the length calculation.

Example 1:

```Input:

1
/   \
3     2
/ \     \
5   3     9

Output: 4
Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).
```

Example 2:

```Input:

1
/
3
/ \
5   3

Output: 2
Explanation: The maximum width existing in the third level with the length 2 (5,3).
```

Example 3:

```Input:

1
/ \
3   2
/
5

Output: 2
Explanation: The maximum width existing in the second level with the length 2 (3,2).
```

Example 4:

```Input:

1
/ \
3   2
/     \
5       9
/         \
6           7
Output: 8
Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).

```

Note: Answer will in the range of 32-bit signed integer.

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#### Approach Framework

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Explanation

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As we need to reach every node in the given tree, we will have to traverse the tree, either with a depth-first search, or with a breadth-first search.

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The main idea in this question is to give each node a `position` value. If we go down the left neighbor, then `position -> position * 2`; and if we go down the right neighbor, then `position -> position * 2 + 1`. This makes it so that when we look at the position values `L` and `R` of two nodes with the same depth, the width will be `R - L + 1`.

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#### Approach #1: Breadth-First Search [Accepted]

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Intuition and Algorithm

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Traverse each node in breadth-first order, keeping track of that node\'s position. For each depth, the first node reached is the left-most, while the last node reached is the right-most.

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Complexity Analysis

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Time Complexity: where is the number of nodes in the input tree. We traverse every node.

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Space Complexity: , the size of our `queue`.

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#### Approach #2: Depth-First Search [Accepted]

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Intuition and Algorithm

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Traverse each node in depth-first order, keeping track of that node\'s position. For each depth, the position of the first node reached of that depth will be kept in `left[depth]`.

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Then, for each node, a candidate width is `pos - left[depth] + 1`. We take the maximum of the candidate answers.

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Complexity Analysis

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Time Complexity: where is the number of nodes in the input tree. We traverse every node.

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Space Complexity: , the size of the implicit call stack in our DFS.

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Analysis written by: @awice.

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