Formatted question description: https://leetcode.ca/all/1650.html
1650. Lowest Common Ancestor of a Binary Tree III
Level
Medium
Description
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:
class Node {
public int val;
public Node left;
public Node right;
public Node parent;
}
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:
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, 10^5]
. -10^9 <= Node.val <= 10^9
- All
Node.val
are unique. p != q
p
andq
exist in the tree.
Solution
Use a set to store nodes. Start from p
and obtain all the nodes from p
to the root node using the nodes’ parents. Then start from q
and move towards the root. If a node is already in the set, then the node is a common ancestor of p
and q
. The first common ancestor that is in the set is the lowest common ancestor of p
and q
.
/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node parent;
};
*/
class Solution {
public Node lowestCommonAncestor(Node p, Node q) {
Set<Node> set = new HashSet<Node>();
Node temp = p;
while (temp != null) {
set.add(temp);
temp = temp.parent;
}
temp = q;
while (temp != null) {
if (set.contains(temp))
break;
else
temp = temp.parent;
}
return temp;
}
}