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:
[2, 105].-109 <= Node.val <= 109Node.val are unique.p != qp and q will exist in the tree./**
* 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)
return root;
if (root==p || root==q)
return root; // rets that node & its ancestor nodes in subsequent surface runs
// dive in depth time
auto ltNode = lowestCommonAncestor(root->left, p, q);
auto rtNode = lowestCommonAncestor(root->right, p, q);
// float to surface time
if (ltNode && rtNode) // both nodes found
return root;
return ltNode?ltNode:rtNode; // if anyone of them is found & other is not found,
// it is evident that the search was not allowed to
// visit such depth. This indicates only one possibility:
// the node found must have been an ancestor of the one not
// found. Hence, return that found node as LCA
}
};