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).

Note: Answer will in the range of 32-bit signed integer.

class Solution {
    private int max = 1;
        public int widthOfBinaryTree(TreeNode root) {
            if (root == null) return 0;
            List<Integer> startOfLevel = new LinkedList<>();
            helper(root, 0, 1, startOfLevel);
            return max;
        }
        public void helper(TreeNode root, int level, int index, List<Integer> list) {
            if (root == null) return;
            if (level == list.size()){
                 list.add(index);
                 //System.out.println(level+" " +list.toString());
            }
            max = Math.max(max, index + 1 - list.get(level));
            //System.out.println(max);
            helper(root.left, level + 1, index * 2, list);
            helper(root.right, level + 1, index * 2 + 1, list);
        }
}

width of binary tree是指最大的（一层里面最右-最左的index+1）。而二叉树可以描述成数组，节点i的两个孩子的节点分别为2*i和2*i + 1，上面答案的意思是

用一个list把每一层的leftmost的index存起来，具体：如果当前root不为null，而且level == list.size( )，这意味着我们到了新的一层，而且这层的leftmost就是当前节点，我们把它存进list中

然后与当前max作比较并更新，如果是新的max，那一定会是index + 1 - list.get(level)，其中list.get(level)是当前的leftmost的index，由于新传入了index，所以可能会出现新max，我们比较就行了

然后继续向下dfs，root.left的index是当前index*2，root.right的index是当前index * 2 + 1。

这类层层的问题似乎都有list.get(heght), 在level order traversal里变成了res.get(height).add(node)

662. Maximum Width of Binary Tree

标签：not   tin   问题   end   order   func   count   function   NPU   

原文地址：https://www.cnblogs.com/wentiliangkaihua/p/13277332.html