0662. Maximum Width of Binary Tree (M)

Posted 2020-12-16 mapoos

Maximum Width of Binary Tree (M)

题目

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.

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题意

找到给定二叉树的最大宽度，每层的宽度定义为每层第一个非空结点到最后一个非空结点的距离。

思路

层序遍历。建两个队列，一个记录结点，一个记录对应结点在当前层的序号。

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代码实现

Java

class Solution {
    public int widthOfBinaryTree(TreeNode root) {
        Queue<TreeNode> nodes = new LinkedList<>();
        Queue<Integer> order = new LinkedList<>();
        int max = 0;
        if (root != null) {
            nodes.offer(root);
            order.offer(0);
        }
        while (!nodes.isEmpty()) {
            int size = nodes.size();
            int start = 0, end = 0;
            for (int i = 0; i < size; i++) {
                TreeNode cur = nodes.poll();
                int num = order.poll();
                start = i == 0 ? num : start;
                end = i == size - 1 ? num : end;
                if (cur.left != null) {
                    nodes.offer(cur.left);
                    order.offer(num * 2);
                }
                if (cur.right != null) {
                    nodes.offer(cur.right);
                    order.offer(num * 2 + 1);
                }
            }
            max = Math.max(max, end - start + 1);
        }
        return max;
    }
}