958. Check Completeness of a Binary Tree

[namespace-io]

Use BFS to do a level order traversal,
add childrens to the bfs queue,
until we met the first empty node.

For a complete binary tree,
there should not be any node after we met an empty one.

/**
 * 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) {
        queue<TreeNode*> q;

        q.push(root);
        while(!q.empty() && q.front() != nullptr){
            auto front = q.front();
            q.push(front->left);
            q.push(front->right);
            q.pop();
        }
        
        while(!q.empty() && q.front() == nullptr) q.pop();
        
        return q.empty();
    }
};

/**
 * 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) {
        queue<TreeNode*> q;
        
        if(root == nullptr) return true;
        
        bool f = false;
        
        q.push(root);
        
 
        while(!q.empty() && !f){
            for(int i = 0, n = q.size(); i < n; i++){
                auto top = q.front();
                if(!f){
                    if(top->left && top->right){
                        q.push(top->left);
                        q.push(top->right);
                    } else if(!top->left && top->right){
                        return false;
                    } else {
                        if(top->left) q.push(top->left);
                        f = true;
                    }
                } else {
                    if(top->left || top->right) return false;
                }
                
                
                
                q.pop();
            }
        }
        
        for(int i =0, n = q.size(); i < n; i++){
            auto top = q.front();
            if(top->left || top->right) return false;
            q.pop();
        }
        
        return true;
    }
};