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222. Count Complete Tree Nodes

Description

Given the root of a complete binary tree, return the number of the nodes in the tree.

According to Wikipedia, every level, except possibly the last, is completely filled in a complete binary tree, 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.

Design an algorithm that runs in less than O(n) time complexity.

 

Example 1:

Input: root = [1,2,3,4,5,6]
Output: 6

Example 2:

Input: root = []
Output: 0

Example 3:

Input: root = [1]
Output: 1

 

Constraints:

  • The number of nodes in the tree is in the range [0, 5 * 104].
  • 0 <= Node.val <= 5 * 104
  • The tree is guaranteed to be complete.

Solutions

Solution 1: Recursion

We recursively traverse the entire tree and count the number of nodes.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\), where \(n\) is the number of nodes in the tree.

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# 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)

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/**
 * 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);
    }
}

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/**
 * 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);
    }
};

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/**
 * 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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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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/**
 * 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);
};

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/**
 * 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);
    }
}

For this problem, we can also take advantage of the characteristics of a complete binary tree to design a faster algorithm.

Characteristics of a complete binary tree: leaf nodes can only appear on the bottom and second-to-bottom layers, and the leaf nodes on the bottom layer are concentrated on the left side of the tree. It should be noted that a full binary tree is definitely a complete binary tree, but a complete binary tree is not necessarily a full binary tree.

If the number of layers in a full binary tree is \(h\), then the total number of nodes is \(2^h - 1\).

We first count the heights of the left and right subtrees of \(root\), denoted as \(left\) and \(right\).

  1. If \(left = right\), it means that the left subtree is a full binary tree, so the total number of nodes in the left subtree is \(2^{left} - 1\). Plus the \(root\) node, it is \(2^{left}\). Then we recursively count the right subtree.
  2. If \(left > right\), it means that the right subtree is a full binary tree, so the total number of nodes in the right subtree is \(2^{right} - 1\). Plus the \(root\) node, it is \(2^{right}\). Then we recursively count the left subtree.

The time complexity is \(O(\log^2 n)\).

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# 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)

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/**
 * 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;
    }
}

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/**
 * 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;
    }
};

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/**
 * 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
}

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/**
 * 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);
};

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/**
 * 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;
    }
}

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