Question
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).
Solution
TODO
Code
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
ArrayList<int[]> vlevels = new ArrayList<>();
int max = 0;
public int widthOfBinaryTree(TreeNode root) {
dfs(root, 1, 0);
return max;
}
public void dfs(TreeNode root, int vlevel, int hlevel) {
if(root == null) return;
if(hlevel == vlevels.size()) {
vlevels.add(new int[] { vlevel, vlevel });
}
int[] vl = vlevels.get(hlevel);
if(vl[0] > vlevel) vl[0] = vlevel;
if(vl[1] < vlevel) vl[1] = vlevel;
max = Math.max(max, vl[1] - vl[0] + 1);
dfs(root.left, 2*vlevel, hlevel+1);
dfs(root.right, 2*vlevel+1, hlevel+1);
}
}
Performance
O(n)