https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/open in new window
Total Accepted: 67533 Total Submissions: 178900 Difficulty: Easy
题目描述
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]
_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5
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.
Note:
1、 Allofthenodes'valueswillbeunique.；
2、 pandqaredifferentandbothvalueswillexistintheBST.；
题目大意
在一个BST中，查找p和q节点的最小公共祖先。
解题方法
注意是BST，那么使用分而治之的策略，用递归来找到左边和右边的最低的公共祖先。
答案：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root==null||p==root||q==root){
return root;
}
//devide
TreeNode left=lowestCommonAncestor(root.left,p,q);
TreeNode right=lowestCommonAncestor(root.right,p,q);
//conquer
if(left!=null&&right!=null){
return root;
}else if(left!=null){
return left;
}else{
return right;
}
}
}
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AC:11ms二刷，python
第一遍做这个题是两年前，现在用Python刷这个感觉特别简单。
因为BST本身的属性，所以比较节点的值和根节点的值的大小就知道下一步去哪里查找了。很简单，看代码。
递归版本：
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def lowestCommonAncestor(self, root, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
if min(p.val, q.val) <= root.val and max(p.val, q.val) >= root.val:
return root
elif p.val < root.val and q.val < root.val:
return self.lowestCommonAncestor(root.left, p, q)
elif p.val > root.val and q.val > root.val:
return self.lowestCommonAncestor(root.right, p, q)
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循环版本：
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def lowestCommonAncestor(self, root, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
while root:
if p.val < root.val and q.val < root.val:
root = root.left
elif p.val > root.val and q.val > root.val:
root = root.right
else:
break
return root
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这个题是236. Lowest Common Ancestor of a Binary Treeopen in new window的特例，所以可以直接使用236的代码就能通过。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def lowestCommonAncestor(self, root, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
if not root or root == p or root == q:
return root
left = self.lowestCommonAncestor(root.left, p, q)
right = self.lowestCommonAncestor(root.right, p, q)
if left and right:
return root
return left if left else right
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