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235. Lowest Common Ancestor of a Binary Search Tree ​

题目 ​

Given a binary search tree (BST), find the lowest common ancestor (LCA) node 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 and q will exist in the BST.
Related Topics
  • 树
  • 深度优先搜索
  • 二叉搜索树
  • 二叉树

  • 👍 1327
  • 👎 0

    • 思路 ​

    BST 是可以根据所搜节点大小判断搜索方向，因为是自上而下的搜索，寻找在节点在 [p, q] 之间

    超过范围时，可以分别进行left or right 的迭代

    如果在范围内（注意此时是一个左右均闭合的集合） 则可以确定 lowset common ancestor 就是这个节点

    递归解法 ​

    python
    class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        l = min(p.val, q.val)
        r = max(p.val, q.val)

        if not root:
            return

        if l <= root.val <= r:
            return root
        elif root.val < l:
            return self.lowestCommonAncestor(root.right, p, q)
        else:
            return  self.lowestCommonAncestor(root.left, p, q)

    简化后

    python
    class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        if root.val > q.val and root.val > p.val:
            return self.lowestCommonAncestor(root.right, p, q)

        elif root.val < q.val and root.val < p.val:
            return self.lowestCommonAncestor(root.left, p, q)
    
        return root

    复杂度分析 ​

    • 时间复杂度 O(N)
      • 极端情况下 tree 变成 linkedList 模式
    • 空间复杂度 O(N)

    迭代法 ​

    只需要找到目标范围的节点即可结束搜索，可以使用迭代法进行节点寻找，可将空间复杂度降低为1

    py
    # leetcode submit region begin(Prohibit modification and deletion)
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        node = root

        while node:
            if node.val > p.val and node.val > q.val:
                node = node.left
            elif node.val < p.val and node.val < q.val:
                node = node.right
            else:
                return node

        return
        
# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(N)
    • 空间复杂度 O(1)