# 958.Check-Completeness-of-a-Binary-Tree

## 题目地址

https://leetcode.com/problems/check-completeness-of-a-binary-tree/

## 题目描述

``````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.``````

## 代码

### Approach #1 BFS

``````/**
* 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) {
List<ANode> nodes = new ArrayList();
int i = 0;
while (i < nodes.size()) {
ANode anode = nodes.get(i++);
if (anode.node != null) {
nodes.add(new ANode(anode.node.right, anode.code * 2 + 1))；
}
}

return nodes.get(i - 1).code == nodes.size();
}
}

class ANode {
TreeNode node;
int code;
ANode(TreeNode node, int code) {
this.node = node;
this.code = code;
}
}``````

### Approach #2 BFS

`Peek() != null`

``````class Solution {
public boolean isCompleteTree(TreeNode root) {
bfs.offer(root);
while (bfs.peek() != null) {
TreeNode node = bfs.poll();
bfs.offer(node.left);
bfs.offer(node.right);
}
while (!bfs.isEmpty() && bfs.peek() == null)
bfs.poll();
return bfs.isEmpty();
}
}``````

Last updated