problem

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).”

Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]

Example 1:

1
2
3
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:

1
2
3
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.

Note:

All of the nodes’ values will be unique. p and q are different and both values will exist in the BST.

初步分析: 1.分别得到两个节点的路径,再比较路径

approach 1:递归

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
/*
* @lc app=leetcode id=235 lang=java
*
* [235] Lowest Common Ancestor of a Binary Search Tree
*/
/**
* 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 || p == null || q == null) {
return null;
}

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p和q都大于root,去左子树找;如果p和q都小于root,去右子树找;否则,p和q进行分叉,root就是要找的节点

非递归版本:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
while (root != null) {
if (root.val > p.val && root.val > q.val) {
root = root.left;
} else if (root.val < p.val && root.val < q.val) {
root = root.right;
} else {
break;
}
}
return root;
}
}

复杂度

  • time:O(n)
  • space:O(n)

  tree

Comments

Your browser is out-of-date!

Update your browser to view this website correctly. Update my browser now

×