LeetCode - Maximum Width of Binary Tree

Question Definition

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.

Java Solution

public int widthOfBinaryTree(TreeNode root) {
    if(root == null) return 0;
    Queue<TreeNode> queue = new LinkedList<>();
    root.val = 1;
    queue.add(root);
    queue.add(null);
    int max = 1;
    int start = -1;
    while(!queue.isEmpty()){
        TreeNode node = queue.poll();
        if(node == null){
            start = -1;
            if(!queue.isEmpty())
                queue.add(null);
        }else{
            if(node.left != null){
                node.left.val = node.val * 2 - 1;
                queue.add(node.left);
                if(start == -1)
                    start = node.left.val;
                else
                    max = Math.max(max, node.left.val - start + 1);
            }
            if(node.right != null){
                node.right.val = node.val * 2;
                queue.add(node.right);
                if(start == -1)
                    start = node.right.val;
                else
                    max = Math.max(max, node.right.val - start + 1);
            }
        }
    }
    return max;
}

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