Maximum Width of Binary Tree v2

 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).

Solution: 

public int widthOfBinaryTree(TreeNode root) {
        if (root == null) {
            return 0;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        int max = 1, leftmost = 1, rightmost = 1;
        root.val = 1;
        queue.add(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.remove();
                if (i == 0) {
                    leftmost = cur.val;
                }
                if (i == size - 1) {
                    rightmost = cur.val;
                }
                if (cur.left != null) {
                    cur.left.val = cur.val * 2;
                    queue.add(cur.left);
                }
                if (cur.right != null) {
                    cur.right.val = cur.val * 2 + 1;
                    queue.add(cur.right);
                }
            }
            max = Math.max(max, rightmost - leftmost + 1);

        }
        return max;
    }