[leetcode] Lowest Common Ancestor of a Binary Search Tree


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.

  • if p->val > q->val, swap q and p
  • if root == null, reutrn null
  • if root->val >= p->val && root->val <= q->val return root; //root is the lowest common ancestor
  • if root->val < p->val , recursively find in root->right
  • else , recursively find in root->left
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 //duplicates in tree?
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(p->val > q->val){
            TreeNode * tmp = p;
            p = q;
            q = tmp;
        }
        return _lowestCommonAncestor(root, p, q);
    }
    TreeNode * _lowestCommonAncestor(TreeNode * root, TreeNode * p, TreeNode * q){
        if(root == nullptr){
            return nullptr;
        }
        else if(p->val <= root->val && q->val >= root->val){
            return root;//find the common ancestor
        }
        else if(p->val > root->val){
            return _lowestCommonAncestor(root->right, p, q);
        }
        else{
            return _lowestCommonAncestor(root->left, p, q);
        }
        
    }
};

 

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