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"content": "<p>Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.</p>\n\n<p>According to the <a href=\"https://en.wikipedia.org/wiki/Lowest_common_ancestor\" target=\"_blank\">definition of LCA on Wikipedia</a>: “The lowest common ancestor is defined between two nodes <code>p</code> and <code>q</code> as the lowest node in <code>T</code> that has both <code>p</code> and <code>q</code> as descendants (where we allow <b>a node to be a descendant of itself</b>).”</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png\" style=\"width: 200px; height: 190px;\" />\n<pre>\n<strong>Input:</strong> root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8\n<strong>Output:</strong> 6\n<strong>Explanation:</strong> The LCA of nodes 2 and 8 is 6.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png\" style=\"width: 200px; height: 190px;\" />\n<pre>\n<strong>Input:</strong> root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> root = [2,1], p = 2, q = 1\n<strong>Output:</strong> 2\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li>The number of nodes in the tree is in the range <code>[2, 10<sup>5</sup>]</code>.</li>\n\t<li><code>-10<sup>9</sup> <= Node.val <= 10<sup>9</sup></code></li>\n\t<li>All <code>Node.val</code> are <strong>unique</strong>.</li>\n\t<li><code>p != q</code></li>\n\t<li><code>p</code> and <code>q</code> will exist in the BST.</li>\n</ul>\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(int x) : val(x), left(NULL), right(NULL) {}\n * };\n */\n\nclass Solution {\npublic:\n TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {\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(int x) { val = x; }\n * }\n */\n\nclass Solution {\n public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {\n \n }\n}",
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"code": "# Definition for a binary tree node.\n# class TreeNode(object):\n# def __init__(self, x):\n# self.val = x\n# self.left = None\n# self.right = None\n\nclass Solution(object):\n def lowestCommonAncestor(self, root, p, q):\n \"\"\"\n :type root: TreeNode\n :type p: TreeNode\n :type q: TreeNode\n :rtype: TreeNode\n \"\"\"\n ",
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