算法leetcode222. 完全二叉树的节点个数（rust和go）

Posted 2022-09-15 二当家的白帽子

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文章目录

• 222. 完全二叉树的节点个数：
• 样例 1：
• 样例 2：
• 样例 3：
• 提示：
• 分析
• 题解
• • rust
  • go
  • typescript
  • python
  • c
  • c++
  • java
• 原题传送门：https://leetcode.cn/problems/count-complete-tree-nodes/

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222. 完全二叉树的节点个数：

给你一棵 完全二叉树 的根节点 root ，求出该树的节点个数。

完全二叉树 的定义如下：在完全二叉树中，除了最底层节点可能没填满外，其余每层节点数都达到最大值，并且最下面一层的节点都集中在该层最左边的若干位置。若最底层为第 h 层，则该层包含 1~ 2^h 个节点。

样例 1：

输入：
	root = [1,2,3,4,5,6]
	
输出：
	6

样例 2：

输入：
	root = []
	
输出：
	0

样例 3：

输入：
	root = [1]
	
输出：
	1

提示：

• 树中节点的数目范围是[0, 5 * 10⁴]
• 0 <= Node.val <= 5 * 10⁴
• 题目数据保证输入的树是 完全二叉树

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分析

• 面对这道算法题目，二当家的陷入了沉思。
• 直接用递归套娃大法即可。
• 但是题目一再强调是完全二叉树，用递归套娃大法并没有考虑这个条件。
• 其实我们只需要看最底层叶子层的情况，找到叶子节点存在和不存在的临界点。

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题解

rust

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode 
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// 
//
// impl TreeNode 
//   #[inline]
//   pub fn new(val: i32) -> Self 
//     TreeNode 
//       val,
//       left: None,
//       right: None
//     
//   
// 
use std::rc::Rc;
use std::cell::RefCell;
impl Solution 
    pub fn count_nodes(root: Option<Rc<RefCell<TreeNode>>>) -> i32 
        if root.is_none() 
            return 0;
        
        let root = root.as_ref().unwrap().borrow();
        return 1 + Solution::count_nodes(root.left.clone()) + Solution::count_nodes(root.right.clone());
    

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go

/**
 * Definition for a binary tree node.
 * type TreeNode struct 
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * 
 */
func countNodes(root *TreeNode) int 
    if root == nil 
		return 0
	
	return 1 + countNodes(root.Left) + countNodes(root.Right)

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typescript

/**
 * Definition for a binary tree node.
 * class TreeNode 
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) 
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     
 * 
 */

function countNodes(root: TreeNode | null): number 
    if (!root) 
		return 0;
	
	return 1+ countNodes(root.left) + countNodes(root.right);
;

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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 countNodes(self, root: TreeNode) -> int:
        if root is None:
            return 0
        return 1 + self.countNodes(root.left) + self.countNodes(root.right)

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c

/**
 * Definition for a binary tree node.
 * struct TreeNode 
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * ;
 */


int countNodes(struct TreeNode* root)
    if (root == NULL) 
        return 0;
    
    return 1 + countNodes(root->left) + countNodes(root->right);

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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:
    int countNodes(TreeNode* root) 
        if (root == NULL) 
            return 0;
        
        return 1 + countNodes(root->left) + countNodes(root->right);
    
;

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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 int countNodes(TreeNode root) 
        if (root == null) 
            return 0;
        
        return 1 + countNodes(root.left) + countNodes(root.right);
    

/**
 * 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 int countNodes(TreeNode root) 
        if (root == null) 
            return 0;
        
        
        // 取得二叉树层级
        int level = 0;
        TreeNode node = root;
        while (node.left != null) 
            level++;
            node = node.left;
        
        
        // 二分查找叶子节点存在和不存在的临界点
        int low = 1 << level, high = (1 << (level + 1)) - 1, bits = 1 << (level - 1);
        while (low < high) 
            int mid = (high + low + 1) / 2;
            if (exists(root, bits, mid)) 
                low = mid;
             else 
                high = mid - 1;
            
        
        return low;
    

    private boolean exists(TreeNode root, int bits, int k) 
        while (root != null && bits > 0) 
            if ((bits & k) == 0) 
                root = root.left;
             else 
                root = root.right;
            
            bits >>= 1;
        
        return root != null;
    

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原题传送门：https://leetcode.cn/problems/count-complete-tree-nodes/

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