Problem

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.

Solution

class Solution {    public int widthOfBinaryTree(TreeNode root) {        if (root == null) return 0;        return dfs(root, 0, 1, new ArrayList
    
     ());    }    private int dfs(TreeNode root, int level, int id, List
     
       leftnodes) {        if (root == null) return 0;        if (leftnodes.size() <= level) leftnodes.add(id);        int left = dfs(root.left, level+1, id*2, leftnodes);        int right = dfs(root.right, level+1, id*2+1, leftnodes);        return Math.max(id-leftnodes.get(level)+1, Math.max(left, right));    }}