
Problem Statement
In this task, you are presented with a binary tree and are required to determine its maximum width. The maximum width of a binary tree is defined as the greatest number of nodes present between the leftmost and rightmost nodes of any level in the tree, including the null nodes that appear as placeholders in a fully complete version of that level. The answer to this problem is guaranteed to fit within a 32-bit signed integer, avoiding potential overflow issues.
Examples
Example 1
Input:
root = [1,3,2,5,3,null,9]
Output:
4
Explanation:
The maximum width exists in the third level with length 4 (5,3,null,9).
Example 2
Input:
root = [1,3,2,5,null,null,9,6,null,7]
Output:
7
Explanation:
The maximum width exists in the fourth level with length 7 (6,null,null,null,null,null,7).
Example 3
Input:
root = [1,3,2,5]
Output:
2
Explanation:
The maximum width exists in the second level with length 2 (3,2).
Constraints
- The number of nodes in the tree is in the range
[1, 3000]
. -100 <= Node.val <= 100
Approach and Intuition
To tackle this problem effectively, a clear understanding of the definition of width within the context of a binary tree is essential:
- Identify the leftmost and rightmost nodes at each level of the tree. These act as the boundaries for measuring the width of that particular level.
- Calculate the number of positions between these boundary nodes. This number should include all the active nodes and the null nodes that would have been present if the tree was completely filled at that level.
- Iterate through all levels of the tree, compute the width for each, and keep track of the maximum width found.
- Return the maximum width obtained from all levels.
The examples help illustrate the definition and calculation of width:
- In the first example, the maximum width is observed in the third level, where even though one node is null, it is still counted, leading to a width of 4.
- The second example demonstrates that even multiple nulls are counted towards the width if they lie between the leftmost and rightmost non-null nodes.
- The third example shows a scenario with fewer levels, where the maximum width is simply the number of nodes at the widest level, which comprises of two actual nodes, making the width 2.
These examples, in conjunction with the formal definition provided, make it clear that understanding how to traverse levels and count positions is key to solving this problem effectively.
Solutions
- Java
- Python
class Solution {
private Integer maximumWidth = 0;
private HashMap<Integer, Integer> levelMinColIndex;
protected void traverse(TreeNode node, Integer level, Integer columnIndex) {
if (node == null)
return;
if (!levelMinColIndex.containsKey(level)) {
levelMinColIndex.put(level, columnIndex);
}
Integer minIndex = levelMinColIndex.get(level);
maximumWidth = Math.max(maximumWidth, columnIndex - minIndex + 1);
traverse(node.left, level + 1, 2 * columnIndex);
traverse(node.right, level + 1, 2 * columnIndex + 1);
}
public int widthOfBinaryTree(TreeNode root) {
this.levelMinColIndex = new HashMap<Integer, Integer>();
traverse(root, 0, 0);
return this.maximumWidth;
}
}
The provided Java code calculates the maximum width of a binary tree, with the width defined as the number of nodes spanning the maximum horizontal distance between the leftmost and rightmost nodes at any level of the tree. It employs a depth-first search (DFS) traversal approach, utilizing recursion.
Here's the breakdown of how the code operates:
- A private member variable named
maximumWidth
stores the maximum width found during the traversal. - A
HashMap
namedlevelMinColIndex
keeps track of the minimum column index observed for each level, facilitating width calculation by mapping each level to its leftmost node's index.
The traverse
method, which performs the recursive DFS, operates as follows:
- The method terminates if it encounters a null node, effectively handling leaf node children.
- It checks if the current level is already recorded in
levelMinColIndex
. If not, it inserts the current node's column index as the starting index for that level. - It calculates the width for the level by computing the difference between the current node's column index and the minimum index stored in
levelMinColIndex
for that level, updatingmaximumWidth
if the current width exceeds previously recorded values.
The traversal logic is recursive:
- For the left child node, the column index is set to
2 * columnIndex
. - For the right child node, it is
2 * columnIndex + 1
, effectively simulating the binary tree structure's breadth expansion in a linear data structure like an array.
The widthOfBinaryTree
method initializes the necessary structures and kicks off the traversal starting from the root node at column index 0 and level 0.
Upon completion of the traversal, widthOfBinaryTree
returns the calculated maximumWidth
, representing the result for the maximum width of the given binary tree.
This algorithm is efficient in terms of both time and space since it only processes each node once and uses a hashmap to dynamically store the minimum indices per level.
class Solution:
def maxWidthOfTree(self, root: TreeNode) -> int:
# Dictionary to keep track of the first column index at each tree level
level_start_indices = {}
width_max = 0
def traverse_tree(node, lvl, idx):
nonlocal width_max
if node is None:
return
# Store first column index if it's missing for the level
if lvl not in level_start_indices:
level_start_indices[lvl] = idx
# Update the max width using the level's first column index
width_max = max(width_max, idx - level_start_indices[lvl] + 1)
# Depth-first search: Left then Right
traverse_tree(node.left, lvl + 1, 2 * idx)
traverse_tree(node.right, lvl + 1, 2 * idx + 1)
traverse_tree(root, 0, 0)
return width_max
The code provided solves the problem of finding the maximum width of a binary tree using Python. The method maxWidthOfTree
in the Solution
class is implemented to determine this width, where the width is defined as the maximum number of nodes at any level of the tree.
- Initiate a dictionary
level_start_indices
to track the column index of the first node at each level of the tree. - Start with a
width_max
variable set to 0 to track the maximum width encountered. - Define a recursive function
traverse_tree
that traverses the tree nodes using Depth First Search (DFS). This internal function takes the current node, its level (lvl
), and column index (idx
) as arguments:- Returns immediately if the current node is
None
. - Records the column index of the first node at each level in
level_start_indices
if not already set. - Updates
width_max
by comparing it with the current width at that level which is calculated using column indices. - Calls itself to process the left child node with updated level and column index.
- Calls itself to process the right child node with updated level and column index.
- Returns immediately if the current node is
- Kickstarts the traversal from the root node at level 0 and index 0.
- Returns
width_max
which holds the maximum width of the tree.
Use this method to measure the maximum width of a binary tree effectively, ensuring efficient handling with depth-first traversal and careful index management.
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