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222. 完全二叉树的节点个数

English Version

题目描述

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~ 2h 个节点。

 

示例 1：

输入：root = [1,2,3,4,5,6]
输出：6

示例 2：

输入：root = []
输出：0

示例 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

 

进阶：遍历树来统计节点是一种时间复杂度为 O(n) 的简单解决方案。你可以设计一个更快的算法吗？

解法

朴素解法是直接遍历整棵完全二叉树，统计结点个数。时间复杂度 O(n)。

对于此题，我们还可以利用完全二叉树的特点，设计一个更快的算法。

完全二叉树的特点：叶子结点只能出现在最下层和次下层，且最下层的叶子结点集中在树的左部。需要注意的是，满二叉树肯定是完全二叉树，而完全二叉树不一定是满二叉树。

若满二叉树的层数为 h，则总结点数为 2^h - 1。

我们可以先对 root 的左右子树进行高度统计，分别记为 left, right。

  1. 若 left == right，说明左子树是一颗满二叉树。所以左子树的结点总数为 2^left - 1，加上 root 结点，就是 2^left，然后递归统计右子树即可。
  2. 若 left > right，说明右子树是一个满二叉树。所以右子树的结点总数为 2^right - 1，加上 root 结点，就是 2^right，然后递归统计左子树即可。

时间复杂度 O(log²n)。

Python3

# 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:
        def depth(root):
            res = 0
            while root:
                res += 1
                root = root.left
            return res

        if root is None:
            return 0
        left, right = depth(root.left), depth(root.right)
        if left == right:
            return (1 << left) + self.countNodes(root.right)
        return (1 << right) + self.countNodes(root.left)

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;
        }
        int left = depth(root.left);
        int right = depth(root.right);
        if (left == right) {
            return (1 << left) + countNodes(root.right);
        }
        return (1 << right) + countNodes(root.left);
    }

    private int depth(TreeNode root) {
        int res = 0;
        for (; root != null; root = root.left, ++res);
        return res;
    }
}

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) return 0;
        int left = depth(root->left);
        int right = depth(root->right);
        if (left == right) return (1 << left) + countNodes(root->right);
        return (1 << right) + countNodes(root->left);
    }

    int depth(TreeNode* root) {
        int res = 0;
        for (; root != nullptr; ++res, root = root->left);
        return res;
    }
};

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
	}
	depth := func(root *TreeNode) int {
		res := 0
		for root != nil {
			res++
			root = root.Left
		}
		return res
	}
	left, right := depth(root.Left), depth(root.Right)
	if left == right {
		return (1 << left) + countNodes(root.Right)
	}
	return (1 << right) + countNodes(root.Left)
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var countNodes = function (root) {
    if (!root) return 0;
    let depth = function (root) {
        let res = 0;
        for (; root != null; ++res, root = root.left);
        return res;
    };
    const left = depth(root.left);
    const right = depth(root.right);
    if (left == right) {
        return (1 << left) + countNodes(root.right);
    }
    return (1 << right) + countNodes(root.left);
};

C#

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int CountNodes(TreeNode root) {
        if (root == null)
        {
            return 0;
        }
        int left = depth(root.left);
        int right = depth(root.right);
        if (left == right)
        {
            return (1 << left) + CountNodes(root.right);
        }
        return (1 << right) + CountNodes(root.left);
    }

    private int depth(TreeNode root) {
        int res = 0;
        for (; root != null; ++res, root = root.left);
        return res;
    }
}

Rust

use std::cell::RefCell;
use std::rc::Rc;

impl Solution {
    pub fn count_nodes(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        if let Some(node) = root {
            let node = node.borrow();
            let left = Self::depth(&node.left);
            let right = Self::depth(&node.right);
            if left == right {
                Self::count_nodes(node.right.clone()) + (1 << left)
            } else {
                Self::count_nodes(node.left.clone()) + (1 << right)
            }
        } else {
            0
        }
    }

    fn depth(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
        if let Some(node) = root {
            Self::depth(&node.borrow().left) + 1
        } else {
            0
        }
    }
}

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