思路 & 代码

• 应该在二叉树的最近公共祖先之前写= =
• 需要用到二叉搜索树的特性～
  
  

代码 & 思路

• 分成四种情况即可，具体见代码注释
• 时间复杂度 O(logn)

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {// 1. 为p | q的情况if(root == p || root == q){return root;}// 2. p & q 都小于 root，转换在左子树找if(p.val < root.val && q.val < root.val){return lowestCommonAncestor(root.left, p, q);}// 3. 都大于，转换右子树找if(p.val > root.val && q.val > root.val){return lowestCommonAncestor(root.right, p, q);}// 4. 一个大一个小，root 就是答案return root;}
}

• 来个清晰版本的代码

class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if(root == p || root == q){return root;}if(p.val < root.val && q.val < root.val){return lowestCommonAncestor(root.left, p, q);}if(p.val > root.val && q.val > root.val){return lowestCommonAncestor(root.right, p, q);}return root;}
}

二刷

• 注意：与二叉树的最近公共祖先不同，这里可以利用二叉搜索树的性质
• 当然，用二叉树来的方法也可以。

class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if(root == p || root == q) {return root;}if(p.val < root.val && q.val < root.val) {return lowestCommonAncestor(root.left, p, q);}else if(p.val > root.val && q.val > root.val) {return lowestCommonAncestor(root.right, p, q);}return root;}
}