236. Lowest Common Ancestor of a Binary Tree
1. Description
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).”
2. Example
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
3. 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 and q will exist in the tree.
4. Solutions
Depth-First Search && Two Scans
n is number of nodes in the root
Time complexity: O(n)
Space complexity: O(n)
class Solution {
public:
TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *p, TreeNode *q) {
vector<TreeNode *> p_path, q_path;
traverse(root, p, p_path);
traverse(root, q, q_path);
TreeNode *candidate = root;
for (int i = 0, n = min(p_path.size(), q_path.size()); i < n && p_path[i] == q_path[i];
++i) {
candidate = p_path[i];
}
return candidate;
}
private:
bool traverse(TreeNode *root, TreeNode *target, vector<TreeNode *> &path) {
if (root != nullptr) {
path.push_back(root);
if (root == target) {
return true;
}
if (traverse(root->left, target, path) || traverse(root->right, target, path)) {
return true;
}
path.pop_back();
}
return false;
}
};
Depth-First Search && Single Scan
n is number of nodes in the root
Time complexity: O(n)
Space complexity: O(n)
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;
}
};