[LeetCode] 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.

二叉树的最大宽度。题意是给一个结构近似于满二叉树的二叉树，请你返回他的最大宽度是多少。

思路是BFS宽度优先搜索。做法是像level order traversal那样遍历树的每一个节点。在需要一个queue来存储遍历到的node的同时，需要另一个linked list来记录每个node的index。这里有一个结论需要记住，如果一个节点的index为i的话，他的左孩子的index是2 * i，他的右孩子的index是2 * i + 1。这个结论可以帮助规避中间有空节点的情况。

时间O(n)

空间O(n)

Java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int widthOfBinaryTree(TreeNode root) {
        // corner case
        if (root == null) {
            return 0;
        }

        // normal case
        Queue<TreeNode> queue = new LinkedList<>();
        LinkedList<Integer> list = new LinkedList<>();
        int res = 1;
        queue.offer(root);
        list.add(1);
        while (!queue.isEmpty()) {
            int count = queue.size();
            for (int i = 0; i < count; i++) {
                TreeNode cur = queue.poll();
                int curIndex = list.removeFirst();
                if (cur.left != null) {
                    queue.offer(cur.left);
                    list.offer(curIndex * 2);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                    list.offer(curIndex * 2 + 1);
                }
            }
            if (list.size() >= 2) {
                res = Math.max(res, list.peekLast() - list.peekFirst() + 1);
            }
        }
        return res;
    }
}

LeetCode 题目总结