原题链接：

235. 二叉搜索树的最近公共祖先

https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-search-tree/description/

完成情况：

解题思路：

_235二叉搜索树的最近公共祖先_递归

package 代码随想录.树.二叉树;import 代码随想录.树.TreeNode;public class _235二叉搜索树的最近公共祖先_递归 {/**** @param root* @param p* @param q* @return*/public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if (root == null || root == p || root == q){return root;}// 后序遍历TreeNode leftNode = lowestCommonAncestor(root.left, p, q);TreeNode rightNode = lowestCommonAncestor(root.right, p, q);//从下往上找，先到最底层if (leftNode == null && rightNode == null){return null;} else if (leftNode != null && rightNode == null) {return leftNode;} else if (rightNode != null && leftNode == null) {return rightNode;}else { // rightNode != null 并且 leftNode != nullreturn root;}}
}

_235二叉搜索树的最近公共祖先_简洁递归

package 代码随想录.树.二叉树;import 代码随想录.树.TreeNode;public class _235二叉搜索树的最近公共祖先_简洁递归 {/**** @param root* @param p* @param q* @return*/public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if (root.val > p.val && root.val > q.val) {return lowestCommonAncestor(root.left,p,q);}if (root.val < p.val && root.val < q.val){return lowestCommonAncestor(root.right,p,q);}return root;    // 如果当前检测到p，q不在同一侧，那么必然当前root就是最近父节点}
}

参考代码：

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