[LeetCode] 236. Lowest Common Ancestor of a Binary Tree_ Medium tag: DFS, Divide and conquer
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]
Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes5
and1
is3.
Example 2:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes5
and4
is5
, since a node can be a descendant of itself according to the LCA definition.
Note:
- All of the nodes' values will be unique.
- p and q are different and both values will exist in the binary tree.
这个题目还是用Divide and Conquer的思想,然后分别在左右两边tree里面找是否有p,或者g,一旦有了,就分别往上面传,只有在left tree, right tree 各有一个点时,就是lowest common ancestor。
Time:O(n), S: O(n)
Code:
class Solution:
def LCA(self, root, p, g):
if not root or root == p or root == g:
return root
left = self.LCA(root.left, p, g)
right = self.LCA(root.right, p, g)
if left and right:
return root
elif left:
return left
elif right:
return right
return
=> 更简洁
class Solution:
def LCA(self, root, p, g):
if root in [None, p, g]: return root
left, right = self.LCA(root.left, p, g), self.LCA(root.right, p, g)
return root if left and right else left or right
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