958. 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.

题目链接

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

方法思路

Approach1:
思路就是先序遍历得到的是二叉树真实的节点数，而层序遍历并不一定，对比二者就可得出是否为完全二叉树。

class Solution {
    //Runtime: 1 ms, faster than 99.19%
    //Memory Usage: 36.9 MB, less than 57.43%
    int dfs_count;
    public boolean isCompleteTree(TreeNode root) {
        int bfs_count = 0;
        dfs_count = 0;
        bfs_count = levelOrderTraverse(root);
        preOrderTraverse(root);
        return bfs_count == dfs_count ? true : false;
    }
    
    private int levelOrderTraverse(TreeNode root){
        Queue<TreeNode> queue = new LinkedList<>();
        int countNodes = 0;
        queue.offer(root);
        countNodes++;
        while(!queue.isEmpty()){
            TreeNode node = queue.poll();
            if(node.left != null) {
                queue.offer(node.left);
                countNodes++;
            }else 
                return countNodes;
            if(node.right != null) {
                queue.offer(node.right);
                countNodes++;
            }else 
                return countNodes;
        }
        return countNodes;
    }
    
    private void preOrderTraverse(TreeNode root){
        if(root == null) return;
        dfs_count++;
        preOrderTraverse(root.left);
        preOrderTraverse(root.right);
    }
}

Approach2:

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.

class Solution {
    //Runtime: 1 ms, faster than 99.19% 
    //Memory Usage: 36.8 MB, less than 57.43%
    public boolean isCompleteTree(TreeNode root) {
        Queue<TreeNode> bfs = new LinkedList<TreeNode>();
        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();
    }
}