Lowest Common Ancestor of a Binary Tree
Problemβ
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: βThe 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).β
Examplesβ
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
Constraintsβ
- 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
will exist in the tree.
Approachβ
To find the lowest common ancestor in a binary tree, we can use a recursive depth-first search (DFS) approach. The idea is to traverse the tree starting from the root. If we find either of the nodes p
or q
, we return that node. If both nodes are found in different subtrees of a node, then that node is their lowest common ancestor.
Steps:β
- Base Case: If the current node is
null
or matchesp
orq
, return the current node. - Recursive Search:
- Recursively search the left subtree for
p
andq
. - Recursively search the right subtree for
p
andq
.
- Recursively search the left subtree for
- Determine LCA:
- If both left and right recursive calls return non-null, the current node is the LCA.
- If only one of the recursive calls returns non-null, that means both nodes are located in the same subtree, so return the non-null result.
Solutionβ
Javaβ
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || root == p || root == q) {
return root;
}
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
if (left != null && right != null) {
return root;
}
return left != null ? left : right;
}
}
C++β
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (root == nullptr || root == p || root == q) {
return root;
}
TreeNode* left = lowestCommonAncestor(root->left, p, q);
TreeNode* right = lowestCommonAncestor(root->right, p, q);
if (left != nullptr && right != nullptr) {
return root;
}
return left != nullptr ? left : right;
}
};
Pythonβ
# 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 or 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
return left if left else right
Complexity Analysisβ
Time Complexity: O(n)β
Reason: The algorithm visits each node in the tree once, where n is the number of nodes in the tree. Space Complexity: O(h)
Reason: The space complexity is determined by the height h of the tree due to the recursion stack. In the worst case, the height of the tree is O(n) for a skewed tree, but O(log n) for a balanced tree.
Referencesβ
LeetCode Problem: Lowest Common Ancestor of a Binary Tree
Solution Link: LCA Solution on LeetCode Wikipedia Definition: Lowest Common Ancestor