mq042.二叉树的最近公共祖先 Leetcode binaryTree linkedList recursion Dec 09, 2020 题目链接 这个也做过,树是二叉树而非BST所以不能靠数字来判断,可以考虑找到根节点到两个目标节点的路径,然后依次判断从哪个节点开始节点开始不同。这个算法的时间复杂度是O(N),空间复杂度是O(N)。 # 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': path_p, path_q = list(), list() def traversal(node, target, path): if not node: return False path.append(node) if node.val == target: return True if traversal(node.left, target, path) or traversal(node.right, target, path): return True path.pop() traversal(root, p.val, path_p) traversal(root, q.val, path_q) i = 0 while i < len(path_q) and i < len(path_p) and path_p[i] == path_q[i]: res = path_p[i] i += 1 return res PREVIOUSmq040.二叉树的最大路径和NEXTmq060.爬楼梯