236. Lowest Common Ancestor of a Binary Tree
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 , 1 0 5 2, 10^5 2,105].
- − 1 0 9 < = N o d e . v a l < = 1 0 9 -10^9 <= Node.val <= 10^9 −109<=Node.val<=109
- All Node.val are unique.
- p != q
- p and q will exist in the tree.
From: LeetCode
Link: 236. Lowest Common Ancestor of a Binary Tree
Solution:
Ideas:
- Start from the root of the tree.
- If the current node is either p or q, then return the current node. This means that we’ve found one of the two nodes.
- Recursively check the left and right child of the current node.
- If both left and right child recursive calls return non-NULL values, this means both p and q are found in different subtrees. Thus, the current node is their LCA.
- If only one of the left or right child recursive calls returns a non-NULL value, return that non-NULL value. This means one of the nodes is found in a subtree.
Code:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
// Base case: if root is NULL or root is either p or q, return root
if (!root || root == p || root == q) {
return root;
}
// Recursively check left and right subtrees
struct TreeNode* left = lowestCommonAncestor(root->left, p, q);
struct TreeNode* right = lowestCommonAncestor(root->right, p, q);
// If both left and right recursive calls return non-NULL, then root is the LCA
if (left && right) {
return root;
}
// Otherwise, return the non-NULL child
return left ? left : right;
}