LSGO——LeetCode实战（链表系列）：235题 二叉树搜索树的最近祖先(Lowest Common Ancester of a Binary Search Tree)

给定一个二叉搜索树, 找到该树中两个指定节点的最近公共祖先。

百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”

例如，给定如下二叉搜索树: root = [6,2,8,0,4,7,9,null,null,3,5]

示例 1:

输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
输出: 6
解释: 节点 2 和节点 8 的最近公共祖先是 6。
示例 2:

输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
输出: 2
解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-search-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

解法一：

算法：

• 从根节点开始遍历树
• 如果节点 pp 和节点 qq 都在右子树上，那么以右孩子为根节点继续 1 的操作
• 如果节点 pp 和节点 qq 都在左子树上，那么以左孩子为根节点继续 1 的操作
• 如果条件 2 和条件 3 都不成立，这就意味着我们已经找到节 pp 和节点 qq 的 LCA 了

递归

# 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':
       # Value of current node or parent node.
        parent_val = root.val

        # Value of p
        p_val = p.val

        # Value of q
        q_val = q.val

        # If both p and q are greater than parent
        if p_val > parent_val and q_val > parent_val:    
            return self.lowestCommonAncestor(root.right, p, q)
        # If both p and q are lesser than parent
        elif p_val < parent_val and q_val < parent_val:    
            return self.lowestCommonAncestor(root.left, p, q)
        # We have found the split point, i.e. the LCA node.
        else:
            return root

迭代

class Solution:
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """

        # Value of p
        p_val = p.val

        # Value of q
        q_val = q.val

        # Start from the root node of the tree
        node = root

        # Traverse the tree
        while node:

            # Value of current node or parent node.
            parent_val = node.val

            if p_val > parent_val and q_val > parent_val:    
                # If both p and q are greater than parent
                node = node.right
            elif p_val < parent_val and q_val < parent_val:
                # If both p and q are lesser than parent
                node = node.left
            else:
                # We have found the split point, i.e. the LCA node.
                return node

程序三：

# 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':
        while (root.val - p.val) * (root.val - q.val) > 0: root = (root.left, root.right)[p.val > root.val]
        return root