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.
分析:
求出BST的LCA。考虑两种情况,一:p和q分别在root的左右子树中,root即为LCA,二:p和q为上下级关系,此时,递归遍历到root等于p或者等于q时,root即为LCA。
代码:
class Solution(object): def lowestCommonAncestor(self, root, p, q): """ :type root: TreeNode :type p: TreeNode :type q: TreeNode :rtype: TreeNode """ # 如果某一个结点的左右子树中分别包含p,q,那么,该结点就是要找的结点 # 或者,等于号说明,如果p和q是上下级关系的话,那么此时root也为LCA if p.val <= root.val <= q.val or q.val <= root.val <= p.val: return root else: # 否则的话,应该去这个根节点的子树中去查找。 if p.val < root.val: # 去左子树中寻找 return self.lowestCommonAncestor(root.left, p, q) else: # 去右子树中寻找 return self.lowestCommonAncestor(root.right, p, q)