leetcode - 662. Maximum Width of Binary Tree【广度优先遍历 + 层次计数】

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).

答案解析

/**
 * 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) {
    	
    	if(root == null){
    		return 0;
    	}
        Queue<TreeNode> queue =  new LinkedList<>();  //用于树的广度优先遍历。
        Queue<Integer> queuePos =  new LinkedList<>(); //用于保存上面队列中树节点对应的位置标号。
        queue.add(root);
        queuePos.add(1);//在顶层跟结点位置为1.
        
        int countCurrent  = 1;//记录正在遍历的当前层次的剩余数量。
        int countTmp = 0; //记录下一层次结点的数量。
        int max = countCurrent; //记录最大的差距。（目标）（与start 和 end相关）
        int start = 1;//记录某层次结点的最左边的结点。
        int end = 1;//记录某层次结点的最右边的结点。
        
        while(!queue.isEmpty()){
            
        	TreeNode current = queue.poll();
        	end = queuePos.poll();

        	if(current.left != null){
        		queue.add(current.left);
        		queuePos.add(2*end);//分配左孩子结点的序号。
        		countTmp ++;//记录下层结点的数量
        	}
        	if(current.right != null){
        		queue.add(current.right);
        		queuePos.add(2*end +1); //分配右孩子结点的序号
        		countTmp ++;
        	}
            
            //当前层次已遍历完毕，计算max,并且为下一层次的遍历准备。
            if(--countCurrent == 0){
                // 目标比对。
        		if(max < end - start + 1){
        			max = end - start +1;
        		}
        		countCurrent = countTmp;//设置下一层次剩余的数量
        		countTmp = 0;
                //设置下一层结点的start.
        		start = queuePos.isEmpty() ?  1 : queuePos.peek();
        		
        	}
        }
        return max;
    }

}

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