此题前面有类似的题目 不过改为搜索树
/**
* 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) {
if(root == null || root == p || root == q) {
return root;
}
//可以用搜索树的性质直接返回 而不用判断是否为空
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;
}
}