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: 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
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, 105]
.-109 <= Node.val <= 109
All
Node.val
are unique.p != q
p
andq
will exist in the tree.
Solutions
π§ Cpp
/**
* 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:
bool has_element(TreeNode* el, TreeNode* p)
{
if(el == nullptr)
return false;
if(el == p)
return true;
else
return has_element(el->right, p) || has_element(el->left, p);
}
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)
{
if(!root)
return nullptr;
if (has_element(root, p) && has_element(root, q))
{
if(has_element(root->right, p) && has_element(root->right, q))
return lowestCommonAncestor(root->right, p, q);
else if(has_element(root->left, p) && has_element(root->left, q))
return lowestCommonAncestor(root->left, p, q);
else return root;
}
else
return nullptr;
}
};
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