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.

常规题

/**
* 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 某个结点孩子不全就开始 判断是不是叶子这个过程
queue.add(root);
TreeNode top = null,L = null,R = null;
while(!queue.isEmpty()){
top = queue.poll();
L = top.left; R = top.right;
//第一种情况
if((R != null && L == null))
return false;
//第二种情况 开启了判断叶子的过程 而且又不是叶子 就返回false
if(leaf && (L != null || R != null)) //以后的结点必须是 左右孩子都是null
return false; if(L != null)
queue.add(L); //准确的说是 只要孩子不全就开启leaf,
//但是前面已经否定了有右无左的情况,这里只要判断一下右孩子是不是为空就可以了(如果为空就开启leaf)
if(R != null)
queue.add(R);
else
leaf = true;
}
return true;
}
}

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