/*

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: 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.
*/

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */

type TreeNodeWithNumber struct {
    Node *TreeNode
    Val int
}
func widthOfBinaryTree(root *TreeNode) int {
    if root == nil {
        return 0
    }

    q := []*TreeNodeWithNumber{&TreeNodeWithNumber{root, 1}}
    _max := 0
    for len(q) > 0 {
        _max = max(_max, q[len(q)-1].Val-q[0].Val+1)
        nxt := []*TreeNodeWithNumber{}

        for _, tn := range q {
            t := tn.Node
            v := tn.Val
            if t.Left != nil {
                nxt = append(nxt, &TreeNodeWithNumber{t.Left, v*2})
            }
            if t.Right != nil {
                nxt = append(nxt, &TreeNodeWithNumber{t.Right, v*2+1})
            }
        }
        q = nxt
    }
    return _max
}
func max(a , b int) int {
    if a > b {
        return a
    }
    return b

}