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).”
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 nodes 5 and 1 is 3.
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 nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
Example 3:
Input: root = [1,2], p = 1, q = 2 Output: 1
Constraints:
- The number of nodes in the tree is in the range
[2, 105]. -109 <= Node.val <= 109- All
Node.valare unique. p != qpandqwill exist in the tree.
题目给定一棵二叉树root， 和两个节点p和q，要求找出p和q的最低公共祖先(LCA)。
首先，对于一个节点node，在以下情况下会是节点p和q的最低公共祖先LCA：1）p和q落在节点node的左右子树上；2）节点node就是p和q的其中一个，另一个落在node左子树或右子树上。满足以上两种情况，节点node就是p和q的最低公共祖先。
另外，这题给的限制条件，使本题变得简单了很多，限制条件是：二叉树的节点个数一定大于2且所有节点都是唯一的；给定的两个节点p和q不相等且一定存在。
基本解题思路就是通过递归法遍历二叉树去找节点p和q，但是并需要遍历整棵树，由于题目说了p和q一定存在，这样我们在递归查找只要碰到一个节点等于p或q就可以停止不用继续递归查找该节点左右子树的所有节点，即使是这时另外一个节点还没找到。以下分析一下不需要查找的原因，假设我们碰到了节点p而这时q还没找到，那么q只有两种情况：1）q在p的左子树或右子树里；2）q在p及其左右子树之外的其他地方。如果q在p的左子树或右子树里那么p就是LCA；若果q不在p的左子树或右子树里，那么q一定是在p以上的某个父节点的另外一棵子树里，也就是p和q一定会分别在那个父节点的左右子树里，那么那个父节点就是LCA。
# 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 not root:
return None
if root == p or root == q:
return root
left = self.lowestCommonAncestor(root.left, p, q)
right = self.lowestCommonAncestor(root.right, p, q)
if left and right:
return root
if not left:
return right
return left
