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

**Example 2:**

Input:1 / 3 / \ 5 3Output:2Explanation:The maximum width existing in the third level with the length 2 (5,3).

**Example 3:**

Input:1 / \ 3 2 / 5Output:2Explanation:The maximum width existing in the second level with the length 2 (3,2).

**Example 4:**

Input:1 / \ 3 2 / \ 5 9 / \ 6 7Output:8Explanation: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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\n#### Approach Framework

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

\n

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

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'
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\n**Explanation**

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.

\nThe 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`

.

\n

**Intuition and Algorithm**

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.

\n\n**Complexity Analysis**

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

\n \n - \n
Space Complexity: , the size of our

\n`queue`

. \n

\n

**Intuition and Algorithm**

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]`

.

Then, for each node, a candidate width is `pos - left[depth] + 1`

. We take the maximum of the candidate answers.

**Complexity Analysis**

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

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

\n \n

\n

Analysis written by: @awice.

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