LintCode 88. Lowest Common Ancestor

题目

思路

1.如果root==A或者root==B,返回root作为LCA
2.找到左子树和右子树是否包含A或者B，如果都包含，返回root，否则返回左子树的LCA或者右子树的LCA
3.都不满足，返回None

代码

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""


class Solution:
    """
    @param: root: The root of the binary search tree.
    @param: A: A TreeNode in a Binary.
    @param: B: A TreeNode in a Binary.
    @return: Return the least common ancestor(LCA) of the two nodes.
    """
    def lowestCommonAncestor(self, root, A, B):
        # write your code here
        if not root or root == A or root == B:
            return root
        left = self.lowestCommonAncestor(root.left, A, B)
        right = self.lowestCommonAncestor(root.right, A, B)
        if left and right: return root
        if left: return left
        if right: return right
        return None