Problem
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.
Solution
class Solution {
public int widthOfBinaryTree(TreeNode root) {
if (root == null) return 0;
return dfs(root, 0, 1, new ArrayList<Integer>());
}
private int dfs(TreeNode root, int level, int id, List<Integer> leftnodes) {
if (root == null) return 0;
if (leftnodes.size() <= level) leftnodes.add(id);
int left = dfs(root.left, level+1, id*2, leftnodes);
int right = dfs(root.right, level+1, id*2+1, leftnodes);
return Math.max(id-leftnodes.get(level)+1, Math.max(left, right));
}
}
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