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tags: leetcode
---

# [236. Lowest Common Ancestor of a Binary Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/)
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# My Solution
## The Key Idea for Solving This Coding Question
## C++ Code

/**
 * 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 == nullptr) {
           return nullptr;
        }

        if (root == p || root == q) {
            return root;
        }

        TreeNode *leftSubtree = lowestCommonAncestor(root->left, p, q);
        TreeNode *rightSubtree = lowestCommonAncestor(root->right, p, q);

        if (leftSubtree == nullptr) {
            // Both p and q are not located in the left subtree.
            // Because p and q exist in the binary tree, they are in the
            // right subtree.
            return rightSubtree;
        }

        if (rightSubtree == nullptr) {
            // Both p and q are not located in the right subtree.
            // Because p and q exist in the binary tree, they are in the
            // left subtree.
            return leftSubtree;
        }

        // Both leftSubtree and rightSubtree are not nullptr.
        // root is the lowest common ancestor of p and q.
        return root;

    }
};

## Time Complexity
$O(n)$
* $n$ is the number of nodes in the binary tree.
## Space Complexity
$O(H)$
* $H$ is the height of the binary tree.
* For the best case, the height of the tree is $logn$, which is a balanced binary tree. The best space complexity is $O(logn)$.
* For the worst case, the height of the tree is $n$, which is a skewed binary tree. The worst complexity is $O(n)$.
# Miscellane
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# Test Cases