题目描述补充

给定一个二叉树，编写一个函数来计算给定深度的所有节点值之和。

示例：

给定二叉树：

    3
   / \
  9  20
    /  \
   15   7

• 如果深度为 2，则应返回 9 + 20 = 29。
• 如果深度为 3，则应返回 15 + 7 = 22。

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 sumOfLevels($root, $depth) {
    if ($root == null || $depth <= 0) {
        return 0;
    }
    
    if ($depth == 1) {
        return $root->val;
    }
    
    // 递归计算左右子树对应深度的和
    return sumOfLevels($root->left, $depth - 1) + sumOfLevels($root->right, $depth - 1);
}

// 示例用法
$root = new TreeNode(3);
$root->left = new TreeNode(9);
$root->right = new TreeNode(20);
$root->right->left = new TreeNode(15);
$root->right->right = new TreeNode(7);

echo sumOfLevels($root, 2); // 输出 29
echo "\n";
echo sumOfLevels($root, 3); // 输出 22

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 sumOfLevels(root, depth):
    if not root or depth <= 0:
        return 0
    if depth == 1:
        return root.val
    return sumOfLevels(root.left, depth - 1) + sumOfLevels(root.right, depth - 1)

# 示例用法
root = TreeNode(3)
root.left = TreeNode(9)
root.right = TreeNode(20)
root.right.left = TreeNode(15)
root.right.right = TreeNode(7)

print(sumOfLevels(root, 2))  # 输出 29
print(sumOfLevels(root, 3))  # 输出 22

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 sumOfLevels(root, depth) {
    if (!root || depth <= 0) {
        return 0;
    }
    if (depth === 1) {
        return root.val;
    }
    return sumOfLevels(root.left, depth - 1) + sumOfLevels(root.right, depth - 1);
}

// 示例用法
const root = new TreeNode(3);
root.left = new TreeNode(9);
root.right = new TreeNode(20);
root.right.left = new TreeNode(15);
root.right.right = new TreeNode(7);

console.log(sumOfLevels(root, 2)); // 输出 29
console.log(sumOfLevels(root, 3)); // 输出 22

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