3Etpl5.json

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        "question": {
            "questionId": "1000306",
            "questionFrontendId": "LCR 049",
            "categoryTitle": "Algorithms",
            "boundTopicId": 910300,
            "title": "求根节点到叶节点数字之和",
            "titleSlug": "3Etpl5",
            "content": "<p>English description is not available for the problem. Please switch to Chinese.</p>\n",
            "translatedTitle": "求根节点到叶节点数字之和",
            "translatedContent": "<p>给定一个二叉树的根节点 <code>root</code> ，树中每个节点都存放有一个 <code>0</code> 到 <code>9</code> 之间的数字。</p>\n\n<div class=\"original__bRMd\">\n<div>\n<p>每条从根节点到叶节点的路径都代表一个数字：</p>\n\n<ul>\n\t<li>例如，从根节点到叶节点的路径 <code>1 -&gt; 2 -&gt; 3</code> 表示数字 <code>123</code> 。</li>\n</ul>\n\n<p>计算从根节点到叶节点生成的 <strong>所有数字之和</strong> 。</p>\n\n<p><strong>叶节点</strong> 是指没有子节点的节点。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/02/19/num1tree.jpg\" style=\"width: 212px; height: 182px;\" />\n<pre>\n<strong>输入：</strong>root = [1,2,3]\n<strong>输出：</strong>25\n<strong>解释：</strong>\n从根到叶子节点路径 <code>1-&gt;2</code> 代表数字 <code>12</code>\n从根到叶子节点路径 <code>1-&gt;3</code> 代表数字 <code>13</code>\n因此，数字总和 = 12 + 13 = <code>25</code></pre>\n\n<p><strong>示例 2：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/02/19/num2tree.jpg\" style=\"width: 292px; height: 302px;\" />\n<pre>\n<strong>输入：</strong>root = [4,9,0,5,1]\n<strong>输出：</strong>1026\n<strong>解释：</strong>\n从根到叶子节点路径 <code>4-&gt;9-&gt;5</code> 代表数字 495\n从根到叶子节点路径 <code>4-&gt;9-&gt;1</code> 代表数字 491\n从根到叶子节点路径 <code>4-&gt;0</code> 代表数字 40\n因此，数字总和 = 495 + 491 + 40 = <code>1026</code>\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>树中节点的数目在范围 <code>[1, 1000]</code> 内</li>\n\t<li><code>0 &lt;= Node.val &lt;= 9</code></li>\n\t<li>树的深度不超过 <code>10</code></li>\n</ul>\n</div>\n</div>\n\n<p>&nbsp;</p>\n\n<p><meta charset=\"UTF-8\" />注意：本题与主站 129&nbsp;题相同：&nbsp;<a href=\"https://leetcode-cn.com/problems/sum-root-to-leaf-numbers/\">https://leetcode-cn.com/problems/sum-root-to-leaf-numbers/</a></p>\n",
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                    "code": "/**\n * Definition for a binary tree node.\n * struct TreeNode {\n *     int val;\n *     TreeNode *left;\n *     TreeNode *right;\n *     TreeNode() : val(0), left(nullptr), right(nullptr) {}\n *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}\n *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}\n * };\n */\nclass Solution {\npublic:\n    int sumNumbers(TreeNode* root) {\n\n    }\n};",
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                    "code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n *     int val;\n *     TreeNode left;\n *     TreeNode right;\n *     TreeNode() {}\n *     TreeNode(int val) { this.val = val; }\n *     TreeNode(int val, TreeNode left, TreeNode right) {\n *         this.val = val;\n *         this.left = left;\n *         this.right = right;\n *     }\n * }\n */\nclass Solution {\n    public int sumNumbers(TreeNode root) {\n\n    }\n}",
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                    "code": "# Definition for a binary tree node.\n# class TreeNode(object):\n#     def __init__(self, val=0, left=None, right=None):\n#         self.val = val\n#         self.left = left\n#         self.right = right\nclass Solution(object):\n    def sumNumbers(self, root):\n        \"\"\"\n        :type root: TreeNode\n        :rtype: int\n        \"\"\"",
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                    "code": "# Definition for a binary tree node.\n# class TreeNode:\n#     def __init__(self, val=0, left=None, right=None):\n#         self.val = val\n#         self.left = left\n#         self.right = right\nclass Solution:\n    def sumNumbers(self, root: TreeNode) -> int:",
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                    "code": "/**\n * Definition for a binary tree node.\n * struct TreeNode {\n *     int val;\n *     struct TreeNode *left;\n *     struct TreeNode *right;\n * };\n */\n\n\nint sumNumbers(struct TreeNode* root){\n\n}",
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                    "code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n *     public int val;\n *     public TreeNode left;\n *     public TreeNode right;\n *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {\n *         this.val = val;\n *         this.left = left;\n *         this.right = right;\n *     }\n * }\n */\npublic class Solution {\n    public int SumNumbers(TreeNode root) {\n\n    }\n}",
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                    "code": "/**\n * Definition for a binary tree node.\n * function TreeNode(val, left, right) {\n *     this.val = (val===undefined ? 0 : val)\n *     this.left = (left===undefined ? null : left)\n *     this.right = (right===undefined ? null : right)\n * }\n */\n/**\n * @param {TreeNode} root\n * @return {number}\n */\nvar sumNumbers = function(root) {\n\n};",
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                    "code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n *     val: number\n *     left: TreeNode | null\n *     right: TreeNode | null\n *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {\n *         this.val = (val===undefined ? 0 : val)\n *         this.left = (left===undefined ? null : left)\n *         this.right = (right===undefined ? null : right)\n *     }\n * }\n */\n\nfunction sumNumbers(root: TreeNode | null): number {\n\n};",
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                    "code": "/**\n * Definition for a binary tree node.\n * type TreeNode struct {\n *     Val int\n *     Left *TreeNode\n *     Right *TreeNode\n * }\n */\nfunc sumNumbers(root *TreeNode) int {\n\n}",
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                    "code": "/**\n * Definition for a binary tree node.\n * class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) {\n *   var value: Int = _value\n *   var left: TreeNode = _left\n *   var right: TreeNode = _right\n * }\n */\nobject Solution {\n    def sumNumbers(root: TreeNode): Int = {\n\n    }\n}",
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