662. Maximum Width of Binary Tree

go-leetcode-662.go

/*
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.

When seeing "full binary tree", I though of traits of binary tree.
For example, 2^h-1 nodes for perfect binary tree and each level has 2^(h-1) nodes

So following the idea...Let's do implement, and the below implementation took 8ms only at top 100%
*/

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

func widthOfBinaryTree(root *TreeNode) int {
    if root == nil {
        return 0
    }
    
    q := make([]*Width,0)
    max := 0
    q = append(q, &Width{root, 1})
    for len(q) > 0 {
        n := len(q)
        _max := q[n-1].W-q[0].W+1
        if max < _max {
            max = _max
        }
        nxt := make([]*Width,0)
        for _,w := range q {
            v := w.W
            if w.T.Left != nil {
                nxt = append(nxt,&Width{w.T.Left,v*2})
            }
            if w.T.Right != nil {
                nxt = append(nxt,&Width{w.T.Right,v*2+1})
            }
        }
        q = nxt
    }
    return max
}