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
\nAs 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
.
Intuition and Algorithm
\nTraverse 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\nComplexity Analysis
\nTime Complexity: where is the number of nodes in the input tree. We traverse every node.
\nSpace Complexity: , the size of our queue
.
Intuition and Algorithm
\nTraverse 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
\nTime Complexity: where is the number of nodes in the input tree. We traverse every node.
\nSpace Complexity: , the size of the implicit call stack in our DFS.
\nAnalysis written by: @awice.
\n