Leetcode1676. Lowest Common Ancestor of a Binary Tree IV [Python]
2022/1/28 17:04:18
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初步的思路是把长度超过2的node做2分,划分到长度为1 或者2的node sublist,这样就可以拿到LCA里处理。但是这样做会在第54(/57)个TC处TLE。先把这个写法留下,之后写可以全部过的版本。
class Solution: def lowestCommonAncestor(self, root: 'TreeNode', nodes: 'List[TreeNode]') -> 'TreeNode': if len(nodes) == 1:return nodes[0] if len(nodes) == 2:return self.findLCA(root, nodes[0], nodes[1]) if len(nodes) > 2: sep = len(nodes)//2 nodelist1 = nodes[:sep] nodelist2 = nodes[sep:] ans1 = self.lowestCommonAncestor(root, nodelist1) ans2 = self.lowestCommonAncestor(root, nodelist2) return self.findLCA(root, ans1, ans2) def findLCA(self, root, node1, node2): if not root:return None if root.val == node1.val or root.val == node2.val:return root leftpath = self.findLCA(root.left, node1, node2) rightpath = self.findLCA(root.right, node1, node2) if leftpath and rightpath:return root return leftpath if leftpath else rightpath
接下来是全部通过的版本,其实只要这样思考,root在nodes中,那一定是nodes, 而不在,则尝试root.left 和root.right看是否在nodes中,在就返回,不在,则继续node的left和right的left和right孙节点是否在nodes中,直到找到。
class Solution: def lowestCommonAncestor(self, root: 'TreeNode', nodes): if not root:return None if root in nodes:return root left = self.lowestCommonAncestor(root.left, nodes) right = self.lowestCommonAncestor(root.right, nodes) if left and right:return root if left:return left if right:return right return None
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