235. Lowest Common Ancestor of a Binary Search Tree [easy] (Python)

题目链接

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

题目原文

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.

题目翻译

思路方法

思路一

根据二叉搜索树的性质，对于树中从root开始的节点：
如果p和q的值如果都小于root的值，那么它们的最低公共祖先一定在root的左子树；如果p和q的值如果都大于root的值，那么它们的最低公共祖先一定在root的右子树；其他情况则说明最低公共祖先就是root节点。如此循环判断即可。

代码

# 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
        """
        pointer = root
        while pointer:
            if p.val > pointer.val and q.val > pointer.val:
                pointer = pointer.right
            elif p.val < pointer.val and q.val < pointer.val:
                pointer = pointer.left
            else:
                return pointer

思路二

类似前面的思路，不过是用递归实现。

代码

# 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:
            return None
        if 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)
        else:
            return root

思路三

更通用的做法是，先分别找到从根节点到两个节点的路径，根据这两个路径找最低公共祖先。方便的是，对于二叉搜索数，不需要BFS或DFS遍历节点找路径。

代码

# 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
        """
        pathp = self.findPath(root, p)
        pathq = self.findPath(root, q)
        res = root
        for i in xrange(1, min(len(pathp), len(pathq))):
            if pathp[i] == pathq[i]:
                res = pathp[i]
        return res


    def findPath(self, root, p):
        path = []
        while root.val != p.val:
            path.append(root)
            if p.val > root.val:
                root = root.right
            elif p.val < root.val:
                root = root.left
        path.append(p)
        return path

PS: 新手刷LeetCode，新手写博客，写错了或者写的不清楚还请帮忙指出，谢谢！
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