Lowest Common Ancestor of a Binary Search Tree

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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

递归解法

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param {TreeNode} root
    # @param {TreeNode} p
    # @param {TreeNode} q
    # @return {TreeNode}
    def lowestCommonAncestor(self, root, p, q):
        if not root or not p or not q:
            return None
        if (max(p.val, q.val) < root.val):
            return self.lowestCommonAncestor(root.left, p, q)
        elif (min(p.val, q.val) > root.val):
            return self.lowestCommonAncestor(root.right, p, q)
        else:
            return root

非递归解法

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param {TreeNode} root
    # @param {TreeNode} p
    # @param {TreeNode} q
    # @return {TreeNode}
    def lowestCommonAncestor(self, root, p, q):
        while root:
            if max(p.val, q.val) < root.val:
                root = root.left
            elif min(p.val, q.val) > root.val:
                root = root.right
            else:
                return root
        return None