LeetCode 235. Lowest Common Ancestor of a Binary Search Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

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).”

_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5
For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

/**
 * 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==NULL || p==NULL || q==NULL) 
            return NULL;
        if((p->val >= root->val && q->val <= root ->val) || (p->val <= root->val && q->val >= root->val)) {
            return root;
        }
        if(p->val > root->val && q->val > root->val) {
            return lowestCommonAncestor(root->right, p, q);
        }
        if(p->val < root->val && q->val < root->val) {
            return lowestCommonAncestor(root->left, p, q);
        }
    }
};

/**

* 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==NULL || p==NULL || q==NULL)

return NULL;

if((p->val >= root->val && q->val <= root ->val) || (p->val <= root->val && q->val >= root->val)) {

return root;

}

if(p->val > root->val && q->val > root->val) {

return lowestCommonAncestor(root->right, p, q);

}

if(p->val < root->val && q->val < root->val) {

return lowestCommonAncestor(root->left, p, q);

}

};

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