662. Maximum Width of Binary Tree

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: 4Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).

Example 2:

Input:           1         /          3           / \             5   3     Output: 2Explanation: The maximum width existing in the third level with the length 2 (5,3).

Example 3:

Input:           1         / \        3   2        /              5      Output: 2Explanation: The maximum width existing in the second level with the length 2 (3,2).

Example 4:

Input:           1         / \        3   2       /     \        5       9      /         \    6           7Output: 8Explanation: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.

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */class Solution {    public int widthOfBinaryTree(TreeNode root) {        if (root == null)return 0;int re = 1;Queue<TreeNode> data = new LinkedList<>();data.add(root);while (!data.isEmpty()) {int n = data.size();int cur = 0;int count = 0;while (n-- > 0) {TreeNode tn = data.poll();if (tn.left != null) {if (count != 0) {count = count + cur + 1;while (cur-- > 0)data.add(new TreeNode(-1));} elsecount += 1;cur = 0;data.add(tn.left);} else {cur++;}if (tn.right != null) {if (count != 0) {count = count + cur + 1;while (cur-- > 0)data.add(new TreeNode(-1));} elsecount += 1;cur = 0;data.add(tn.right);} else {cur++;}}re = re >= count ? re : count;}return re;    }}