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ylb 提交于 2024-02-21 08:52 . feat: add problem tags (#2361)

958. 二叉树的完全性检验

English Version

题目描述

给你一棵二叉树的根节点 root ,请你判断这棵树是否是一棵 完全二叉树 。

在一棵 完全二叉树 中,除了最后一层外,所有层都被完全填满,并且最后一层中的所有节点都尽可能靠左。最后一层(第 h 层)中可以包含 1 到 2h 个节点。

 

示例 1:

输入:root = [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,节点值为 {1} 和 {2,3} 的两层),且最后一层中的所有节点({4,5,6})尽可能靠左。

示例 2:

输入:root = [1,2,3,4,5,null,7]
输出:false
解释:值为 7 的节点不满足条件「节点尽可能靠左」。

 

提示:

  • 树中节点数目在范围 [1, 100]
  • 1 <= Node.val <= 1000

解法

方法一

# 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:
        q = deque([root])
        while q:
            node = q.popleft()
            if node is None:
                break
            q.append(node.left)
            q.append(node.right)
        return all(node is None for node in q)
/**
 * 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) {
        Deque<TreeNode> q = new LinkedList<>();
        q.offer(root);
        while (q.peek() != null) {
            TreeNode node = q.poll();
            q.offer(node.left);
            q.offer(node.right);
        }
        while (!q.isEmpty() && q.peek() == null) {
            q.poll();
        }
        return q.isEmpty();
    }
}
/**
 * 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{{root}};
        while (q.front()) {
            root = q.front();
            q.pop();
            q.push(root->left);
            q.push(root->right);
        }
        while (!q.empty() && !q.front()) q.pop();
        return q.empty();
    }
};
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isCompleteTree(root *TreeNode) bool {
	q := []*TreeNode{root}
	for q[0] != nil {
		root = q[0]
		q = q[1:]
		q = append(q, root.Left)
		q = append(q, root.Right)
	}
	for len(q) > 0 && q[0] == nil {
		q = q[1:]
	}
	return len(q) == 0
}
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