题目描述补充

给定一棵二叉树，请你编写一个函数来计算这棵树所有叶节点的和（叶节点是指没有子节点的节点）。

示例

考虑以下二叉树：

    3
   / \
  9  20
    /  \
   15   7

这棵树的叶节点为 9、15 和 7，所以它们的和为 31。

PHP 示例代码

<?php

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 sumOfLeafNodes($root) {
    if ($root === null) {
        return 0;
    }
    
    if ($root->left === null && $root->right === null) {
        // 当前节点是叶节点
        return $root->val;
    }
    
    // 递归计算左子树和右子树的叶节点和
    return sumOfLeafNodes($root->left) + sumOfLeafNodes($root->right);
}

// 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 sumOfLeafNodes($root); // 输出 31

?>

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 sumOfLeafNodes(root):
    if not root:
        return 0
    
    if not root.left and not root.right:
        # 当前节点是叶节点
        return root.val
    
    # 递归计算左子树和右子树的叶节点和
    return sumOfLeafNodes(root.left) + sumOfLeafNodes(root.right)

# 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(sumOfLeafNodes(root))  # 输出 31

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 sumOfLeafNodes(root) {
    if (!root) {
        return 0;
    }
    
    if (!root.left && !root.right) {
        // 当前节点是叶节点
        return root.val;
    }
    
    // 递归计算左子树和右子树的叶节点和
    return sumOfLeafNodes(root.left) + sumOfLeafNodes(root.right);
}

// 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(sumOfLeafNodes(root)); // 输出 31

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