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:
- 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; } }