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: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

TreeNode * LCA(TreeNode * root, TreeNode * p, TreeNode * q)

• if root is null, return null
• if root == p || root == q, return root // find the target node
• else:
  • left = LCA(root->left, p, q)
  • right = LCA(root->right, p, q)
  • if(left && right) return root; //lowest ancestor
  • else return left == nullptr? right: left; //return the non null child

/**
 * 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:
    //what if a node does not exist in tree
    //what if p == nullptr || q == nullptr
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root == nullptr || p == nullptr || q == nullptr){
            return nullptr;
        }
        else if(root == p || root == q){
            return root;
        }
        else{
            //search deeper
            TreeNode * left = lowestCommonAncestor(root->left, p, q);
            TreeNode * right = lowestCommonAncestor(root->right, p, q);
            if((left && right)){
                //lowest ancestor
                return root;
            }
            else{
                //below ancestor or above ancestor
                return left == nullptr? right: left;
            }
        }
    }
};