662. Maximum Width of Binary Tree

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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).

We know that a binary tree can be represented by an array (assume the root begins from the position with index 1 in the array). If the index of a node is i, the indices of its two children are 2*i and 2*i + 1. The idea is to use two arrays (start[] and end[]) to record the the indices of the leftmost node and rightmost node in each level, respectively. For each level of the tree, the width is end[level] - start[level] + 1. Then, we just need to find the maximum width.

Java version:

    public int widthOfBinaryTree(TreeNode root) {
        return dfs(root, 0, 1, new ArrayList<Integer>(), new ArrayList<Integer>());
    }
    
    public int dfs(TreeNode root, int level, int order, List<Integer> start, List<Integer> end){
        if(root == null)return 0;
        if(start.size() == level){
            start.add(order); end.add(order);
        }
        else end.set(level, order);
        int cur = end.get(level) - start.get(level) + 1;
        int left = dfs(root.left, level + 1, 2*order, start, end);
        int right = dfs(root.right, level + 1, 2*order + 1, start, end);
        return Math.max(cur, Math.max(left, right));
    }