题目描述

给定一棵完全二叉树的根节点，请编写一个函数来计算这棵完全二叉树中节点的个数。

注意：

• 完全二叉树的定义是：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，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-1) 个节点。

示例

假设我们有以下完全二叉树：

    1
   / \
  2   3
 / \   \
4   5   6

该完全二叉树有 6 个节点。

PHP 示例代码

class TreeNode {
    public $val;
    public $left;
    public $right;
    
    function __construct($val = 0, $left = null, $right = null) {
        $this->val = $val;
        $this->left = $left;
        $this->right = $right;
    }
}

function countNodes($root) {
    if ($root == null) {
        return 0;
    }
    
    $leftHeight = 0;
    $rightHeight = 0;
    $leftNode = $root->left;
    $rightNode = $root->right;
    
    // 计算左子树的高度
    while ($leftNode != null) {
        $leftHeight++;
        $leftNode = $leftNode->left;
    }
    
    // 计算右子树的高度
    while ($rightNode != null) {
        $rightHeight++;
        $rightNode = $rightNode->right;
    }
    
    // 如果左右子树高度相同，说明左子树是满二叉树
    if ($leftHeight == $rightHeight) {
        return (1 << $leftHeight) - 1 + countNodes($root->right);
    } else {
        // 否则，右子树是满二叉树（少一层）
        return (1 << $rightHeight) - 1 + countNodes($root->left);
    }
}

// 示例用法
$root = new TreeNode(1);
$root->left = new TreeNode(2);
$root->right = new TreeNode(3);
$root->left->left = new TreeNode(4);
$root->left->right = new TreeNode(5);
$root->right->right = new TreeNode(6);

echo countNodes($root); // 输出 6

Python 示例代码

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def countNodes(root):
    if not root:
        return 0
    
    left_height = 0
    right_height = 0
    left_node = root.left
    right_node = root.right
    
    # 计算左子树的高度
    while left_node:
        left_height += 1
        left_node = left_node.left
    
    # 计算右子树的高度
    while right_node:
        right_height += 1
        right_node = right_node.right
    
    # 如果左右子树高度相同，说明左子树是满二叉树
    if left_height == right_height:
        return (1 << left_height) - 1 + countNodes(root.right)
    else:
        # 否则，右子树是满二叉树（少一层）
        return (1 << right_height) - 1 + countNodes(root.left)

# 示例用法
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
root.right.right = TreeNode(6)

print(countNodes(root))  # 输出 6

JavaScript 示例代码

function TreeNode(val, left, right) {
    this.val = (val===undefined ? 0 : val)
    this.left = (left===undefined ? null : left)
    this.right = (right===undefined ? null : right)
}

function countNodes(root) {
    if (!root) {
        return 0;
    }
    
    let leftHeight = 0;
    let rightHeight = 0;
    let leftNode = root.left;
    let rightNode = root.right;
    
    // 计算左子树的高度
    while (leftNode) {
        leftHeight++;
        leftNode = leftNode.left;
    }
    
    // 计算右子树的高度
    while (rightNode) {
        rightHeight++;
        rightNode = rightNode.right;
    }
    
    // 如果左右子树高度相同，说明左子树是满二叉树
    if (leftHeight === rightHeight) {
        return (1 << leftHeight) - 1 + countNodes(root.right);
    } else {
        // 否则，右子树是满二叉树（少一层）
        return (1 << rightHeight) - 1 + countNodes(root.left);
    }
}

// 示例用法
const root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(5);
root.right.right = new TreeNode(6);

console.log(countNodes(root)); // 输出 6

以上代码通过递归和位运算高效地计算了完全二叉树的节点个数。