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:
![](https://spcdn.pages.dev/leetcode/problems/1650.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree%20III/images/binarytree.png)
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:
![](https://spcdn.pages.dev/leetcode/problems/1650.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree%20III/images/binarytree.png)
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.val
are unique. p != q
p
and q
exist in the tree.
Solutions#
Solution 1#
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| """
# Definition for a Node.
class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
self.parent = None
"""
class Solution:
def lowestCommonAncestor(self, p: 'Node', q: 'Node') -> 'Node':
a, b = p, q
while a != b:
a = a.parent if a.parent else q
b = b.parent if b.parent else p
return a
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| /*
// 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) {
Node a = p, b = q;
while (a != b) {
a = a.parent == null ? q : a.parent;
b = b.parent == null ? p : b.parent;
}
return a;
}
}
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| /*
// Definition for a Node.
class Node {
public:
int val;
Node* left;
Node* right;
Node* parent;
};
*/
class Solution {
public:
Node* lowestCommonAncestor(Node* p, Node* q) {
Node* a = p;
Node* b = q;
while (a != b) {
a = a->parent ? a->parent : q;
b = b->parent ? b->parent : p;
}
return a;
}
};
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| /**
* Definition for Node.
* type Node struct {
* Val int
* Left *Node
* Right *Node
* Parent *Node
* }
*/
func lowestCommonAncestor(p *Node, q *Node) *Node {
a, b := p, q
for a != b {
if a.Parent != nil {
a = a.Parent
} else {
a = q
}
if b.Parent != nil {
b = b.Parent
} else {
b = p
}
}
return a
}
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