leetcode之 Lowest Common Ancestor of a Binary Tree

这题是跟Lowest Common Ancestor of a Binary Search Tree相似的思路，不过不再是直接通过比较值来判定，而是要通过看在中序排序中的位置来定。极端的情况比如单链形式的q，p只相差1的情况会非常慢。特意加上了相差1的判定。代码如下：

# 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 __init__(self):
        self.list1 = []
    def inorder(self,root):
        if root.left:
            self.inorder(root.left)
        self.list1.append(root)
        if root.right:
            self.inorder(root.right)
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        self.inorder(root)
        self.indexofroot = self.list1.index(root)
        self.indexofp = self.list1.index(p)
        self.indexofq = self.list1.index(q)
        
        def ancestor(root, p, q):
            if self.indexofq <= self.indexofroot <= self.indexofp or self.indexofq >= self.indexofroot >= self.indexofp:
                return root
            if abs(self.indexofp - self.indexofq) == 1:
                if p.left == q or p.right == q:
                    return p
                else:
                    return p
            elif self.indexofroot < self.indexofp and self.indexofroot < self.indexofq:
                self.indexofroot = self.list1.index(root.right)
                return ancestor(root.right, p, q)
            else:
                self.indexofroot = self.list1.index(root.left)
                return ancestor(root.left, p, q)
        return ancestor(root, p, q)