958 Check Completeness of a Binary Tree 二叉树的完全性检验
Description:
Given the root of a binary tree, determine if it is a complete binary tree.
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:
Example 1:
Input: root = [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: root = [1,2,3,4,5,null,7]
Output: false
Explanation: The node with value 7 isn't as far left as possible.
Constraints:
The number of nodes in the tree is in the range [1, 100].
1 <= Node.val <= 1000
题目描述:
给定一个二叉树，确定它是否是一个完全二叉树。
百度百科中对完全二叉树的定义如下：
若设二叉树的深度为 h，除第 h 层外，其它各层 (1～h-1) 的结点数都达到最大个数，第 h 层所有的结点都连续集中在最左边，这就是完全二叉树。（注：第 h 层可能包含 1~ 2h 个节点。）
示例 :
示例 1：
输入：[1,2,3,4,5,6]
输出：true
解释：最后一层前的每一层都是满的（即，结点值为 {1} 和 {2,3} 的两层），且最后一层中的所有结点（{4,5,6}）都尽可能地向左。
示例 2：
输入：[1,2,3,4,5,null,7]
输出：false
解释：值为 7 的结点没有尽可能靠向左侧。
提示:
树中将会有 1 到 100 个结点。
思路:
层序遍历
遍历到 null 的时候停止遍历
如果还有结点, 说明是非完全二叉树
时间复杂度为 O(n), 空间复杂度为 O(n)
代码:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution
{
public:
bool isCompleteTree(TreeNode* root)
{
queue<TreeNode*> q;
q.push(root);
while (q.front())
{
TreeNode* cur = q.front();
q.pop();
q.push(cur -> left);
q.push(cur -> right);
}
while (!q.empty())
{
if (q.front()) return false;
q.pop();
}
return true;
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean isCompleteTree(TreeNode root) {
Queue<TreeNode> q = new LinkedList<>();
q.add(root);
TreeNode cur;
while ((cur = q.poll()) != null) {
q.add(cur.left);
q.add(cur.right);
}
while (!q.isEmpty()) if (q.poll() != null) return false;
return true;
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def isCompleteTree(self, root: TreeNode) -> bool:
return None not in root._tree_node_to_array()
