思路
由于是BST,思路相比与236题有所简化。
只需要判断2个目标是否都在同一分支即可。
递归实现
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
// recursively
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(p.val > root.val && q.val > root.val)
return lowestCommonAncestor(root.right, p, q);
if(p.val < root.val && q.val < root.val)
return lowestCommonAncestor(root.left, p, q);
return root;
}
}
时间复杂度O(n)
空间复杂度O(n)
迭代实现
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
// iteratively
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
while(root != null) {
if(p.val > root.val && q.val > root.val)
root = root.right;
else if(p.val < root.val && q.val < root.val)
root = root.left;
else
return root;
}
return null;
}
}
时间复杂度O(n)
空间复杂度O(1)