Given a binary tree, determine if it is a complete binary tree.

Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.

Example 1:

Input: [1,2,3,4,5,6]
Output: true
Explanation: Every level before the last is full (ie. levels with node-values {1} and {2, 3}), and all nodes in the last level ({4, 5, 6}) are as far left as possible.

Example 2:

Input: [1,2,3,4,5,null,7]
Output: false
Explanation: The node with value 7 isn't as far left as possible.
 

Note:

1. The tree will have between 1 and 100 nodes.

常規題

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isCompleteTree(TreeNode root) {
        
 
if(root == null)
            return true;
        Queue<TreeNode>queue = new LinkedList<>();
        boolean leaf = false; //如果碰到了 某個結點孩子不全就開始　判斷是不是葉子這個過程
        queue.add(root);
        TreeNode top = null,L = null,R = null;
        while(!queue.isEmpty()){
            top = queue.poll();
            L = top.left;  R = top.right;
            //第一種情況
            if((R != null && L == null))
                return false;
            //第二種情況  開啟了判斷葉子的過程 而且又不是葉子 就返回false
            if(leaf && (L != null || R != null)) //以後的結點必須是 左右孩子都是null
                return false;

            if(L != null)
                queue.add(L);

            //準確的說是　只要孩子不全就開啟leaf，
            //但是前面已經否定了有右無左的情況，這裡只要判斷一下右孩子是不是為空就可以了(如果為空就開啟leaf)
            if(R != null)
                queue.add(R);
            else
                leaf = true;
        }
        return true;
    }
}