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Given two nodes of a binary tree p and q, return their lowest common ancestor (LCA).
Each node will have a reference to its parent node. The definition for Node is below:
<pre>class Node { public int val; public Node left; public Node right; public Node parent; } </pre>
According to the definition of LCA on Wikipedia: "The lowest common ancestor of two nodes p and q in a tree T is the lowest node that has both p and q as descendants (where we allow a node to be a descendant of itself)."
Example 1:

<pre>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. </pre>
Example 2:

<pre>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. </pre>
Example 3:
<pre>Input: root = [1,2], p = 1, q = 2 Output: 1 </pre>
Constraints:
- The number of nodes in the tree is in the range
[2, 10<sup>5</sup>]. -10<sup>9</sup> <= Node.val <= 10<sup>9</sup>- All
Node.valare unique. p != qpandqexist in the tree.
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题目大意:
求带父节点的树中的两个节点的LCA。节点值唯一,且两输入节点不同,且一定存在
解题思路:
N/A
解题步骤:
N/A
注意事项:
- 某一个节点的左右父节点存入set中,另一节点的每个父节点在set中找
Python代码:
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12def lowestCommonAncestor(self, p: 'Node', q: 'Node') -> 'Node':
parent_set = set()
it = p
while it:
parent_set.add(it)
it = it.parent
it = q
while it:
if it in parent_set:
return it
it = it.parent
return None
算法分析:
时间复杂度为O(n + m),空间复杂度O(n),n和m为所有父亲路径长


