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.
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.
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.
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.
Complexity Analysis
Time Complexity: where is the number of nodes in the input tree. We traverse every node.
Space Complexity: , the size of our queue.
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
Time Complexity: where is the number of nodes in the input tree. We traverse every node.
Space Complexity: , the size of the implicit call stack in our DFS.
Analysis written by: @awice.