Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p
and q
as the lowest node in T
that has both p
and q
as descendants (where we allow a node to be a descendant of itself).”
Example 1:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 Output: 6 Explanation: The LCA of nodes 2 and 8 is 6.
Example 2:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4 Output: 2 Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
Example 3:
Input: root = [2,1], p = 2, q = 1 Output: 2
Constraints:
- The number of nodes in the tree is in the range
[2, 105]
. -109 <= Node.val <= 109
- All
Node.val
are unique. p != q
p
andq
will exist in the BST.
二叉搜索树的最小公共祖先。
影子题236。题意是给一个二叉搜索树(BST)和两个节点p和q,请找出两个节点的最小公共祖先。思路是递归,因为BST的性质,所以任何节点的左孩子一定比当前节点小,右孩子一定比当前节点大。明白这一点,代码就不难写。
时间O(n)
空间O(n)
Java实现
1 class Solution { 2 public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { 3 if (root.val > p.val && root.val > q.val) { 4 return lowestCommonAncestor(root.left, p, q); 5 } else if (root.val < p.val && root.val < q.val) { 6 return lowestCommonAncestor(root.right, p, q); 7 } else { 8 return root; 9 } 10 } 11 }
JavaScript实现
1 /** 2 * @param {TreeNode} root 3 * @param {TreeNode} p 4 * @param {TreeNode} q 5 * @return {TreeNode} 6 */ 7 var lowestCommonAncestor = function (root, p, q) { 8 if (root.val > p.val && root.val > q.val) { 9 return lowestCommonAncestor(root.left, p, q); 10 } else if (root.val < p.val && root.val < q.val) { 11 return lowestCommonAncestor(root.right, p, q); 12 } else { 13 return root; 14 } 15 };
相关题目
235. Lowest Common Ancestor of a Binary Search Tree
236. Lowest Common Ancestor of a Binary Tree