Problem description:

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.

Solution:

When level-order traversal in a complete tree, after the last node, all nodes in the queue should be null.
Otherwise, the tree is not complete.

I use the preorder traversal to trace the tree. It is because we need to walk the level from left to right.
example:
we can see that the X is the last node in this level. After X, we’ll go to next level, which should all be NULL.

1
2 3 4 X

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isCompleteTree(TreeNode* root) {
        //idea is to check if there're other node is also empty after we already find an empty node
        queue<TreeNode*> q;
        q.push(root);
        bool end= false;
        while(!q.empty()){
            auto cur= q.front(); q.pop();
            if(!cur)
                end= true;
            else{
                if(end) return false;
                q.push(cur->left);
                q.push(cur->right);
            }
        }
        return true;
    }
};

time complexity: $O(n)$
space complexity: $O(n)$
reference: