Question

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

方法一：递归

递归遍历整棵树，统计结点个数。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为树的结点个数。

# 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: Optional[TreeNode]) -> int:
    if root is None:
      return 0
    return 1 + self.countNodes(root.left) + self.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;
    }
    return 1 + countNodes(root.left) + countNodes(root.right);
  }
}

/**
 * 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;
    }
    return 1 + countNodes(root->left) + countNodes(root->right);
  }
};

/**
 * 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)
}

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 }
  }
}

/**
 * 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;
  }
  return 1 + countNodes(root.left) + countNodes(root.right);
};

/**
 * 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;
    }
    return 1 + CountNodes(root.left) + CountNodes(root.right);
  }
}

方法二：二分查找

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 $left$ 和 $right$。

1. 若 $left = right$，说明左子树是一颗满二叉树，那么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 $left \gt right$，说明右子树是一个满二叉树，那么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 $O(\log^2 n)$。

# 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: Optional[TreeNode]) -> int:
    def depth(root):
      d = 0
      while root:
        d += 1
        root = root.left
      return d

    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)

/**
 * 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 d = 0;
    for (; root != null; root = root.left) {
      ++d;
    }
    return d;
  }
}

/**
 * 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 d = 0;
    for (; root; root = root->left) {
      ++d;
    }
    return d;
  }
};

/**
 * 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
  }
  left, right := depth(root.Left), depth(root.Right)
  if left == right {
    return (1 << left) + countNodes(root.Right)
  }
  return (1 << right) + countNodes(root.Left)
}

func depth(root *TreeNode) (d int) {
  for ; root != nil; root = root.Left {
    d++
  }
  return
}

/**
 * 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) {
  const depth = root => {
    let d = 0;
    for (; root; root = root.left) {
      ++d;
    }
    return d;
  };
  if (!root) {
    return 0;
  }
  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);
};

/**
 * 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 d = 0;
    for (; root != null; root = root.left) {
      ++d;
    }
    return d;
  }
}