LeetCode 236.二叉树的最近公共祖先

给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。

百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”

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

class Solution:
    def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:
        if root is None:
            return None
        if root == p:
            return p 
        if root == q:
            return q
        """
        如果root是null，则说明我们已经找到最底了，返回null表示没找到
        如果root与p相等或者与q相等，则返回root
        如果左子树没找到，递归函数返回null，证明p和q同在root的右侧，那么最终的公共祖先就是右子树找到的结点
        如果右子树没找到，递归函数返回null，证明p和q同在root的左侧，那么最终的公共祖先就是左子树找到的结点

        """
        L = self.lowestCommonAncestor(root.left,p,q)
        R = self.lowestCommonAncestor(root.right,p,q)
        if L and R:
            return root 
        return L if L else R 



class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        has_p,has_q,ancestor = self.solver(root,p,q)
        return ancestor

    def solver(self,root,p,q):
        if root is None:
            return False,False,None 
        has_p_left,has_q_left,ancestor_left = self.solver(root.left,p,q)
        has_p_right,has_q_right,ancestor_right = self.solver(root.right,p,q)
        if has_p_left is True or has_p_right is True or root.val == p.val:
            has_p = True
        else:
            has_p = False
        if has_q_left is True or has_q_right is True or root.val == q.val:
            has_q = True
        else:
            has_q = False
        if ancestor_left is not None:
            ancestor = ancestor_left
        elif ancestor_right is not None:
            ancestor = ancestor_right
        elif ancestor_left is None and ancestor_right is None:
            if has_p is True and has_q is True:
                ancestor = root 
            else: 
                ancestor = None
        return has_p,has_q,ancestor