LeetCode 1644. Lowest Common Ancestor of a Binary Tree II
Tree
Given the root
of a binary tree, return the lowest common ancestor (LCA) of two given nodes, p
and q
. If either node p
or q
does not exist in the tree, return null
. All values of the nodes in the tree are unique.
According to the definition of LCA on Wikipedia: "The lowest common ancestor of two nodes p
and q
in a binary 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)". A descendant of a node x
is a node y
that is on the path from node x
to some leaf node.
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. A node can be a descendant of itself according to the definition of LCA.
Example 3:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 10
Output: null
Explanation: Node 10 does not exist in the tree, so return null.
Constraints:
The number of nodes in the tree is in the range
[1, 104]
.-109 <= Node.val <= 109
All
Node.val
are unique.p != q
Follow up: Can you find the LCA traversing the tree, without checking nodes existence?
Solution
/**
* 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 {
pair<TreeNode*, int> dfs(TreeNode* root, TreeNode* p, TreeNode* q) {
if (root == nullptr) {
return {nullptr, 0};
}
pair<TreeNode*, int> l_res = dfs(root->left, p, q);
pair<TreeNode*, int> r_res = dfs(root->right, p, q);
if (root == p || root == q) {
return {root, 1 + l_res.second + r_res.second};
}
if (l_res.first && r_res.first) {
return {root, 2};
}
return l_res.first ? l_res : r_res;
}
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
pair<TreeNode*, int> res = dfs(root, p, q);
if (res.second < 2) {
return nullptr;
}
return res.first;
}
};
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