Given a binary tree, write a function to get the maximum width of the given tree. The maximum width of a tree is the maximum width among all levels.
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.
It is guaranteed that the answer will in the range of 32-bit signed integer.
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).
Constraints:
- The given binary tree will have between 1 and 3000 nodes.
Notes:
Solution:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
Integer max = 0;
Map<Integer, Integer> leftMostPositions = new HashMap<>();
public int widthOfBinaryTree(TreeNode root) {
dfs(root, 0, 0);
return max;
}
void dfs(TreeNode node, int depth, int position) {
if (node == null) {
return;
}
leftMostPositions.computeIfAbsent(depth, x -> position);
max = Math.max(max, position - leftMostPositions.get(depth) + 1);
dfs(node.left, depth + 1, position * 2);
dfs(node.right, depth + 1, position * 2 + 1);
}
}