(Leet Code c++)Lowest Common Ancestor of a Binary Search Tree

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 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).”

 

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 Output: 6 Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4 Output: 2 Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [2,1], p = 2, q = 1 Output: 2

 

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 and q will exist in the BST.

이진 트리의 최소 공통 부모를 찾는 문제.

백준 알고리즘의 11437 LCA와 같은 문제다.

https://eunchanee.tistory.com/267

(백준 c++)11437 LCA

다만 이 문제의 차이점은 int형이 아닌 TreeNode * 을 리턴해야 하기 때문에,

map을 활용했다.

(노드에 담겨있는 수의 범위가 다르기 때문에)

또한, 노드에 담겨있는 값이 유니크하기 때문에 visited는 없어도 된다.

/**
 * 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:
    map<int, TreeNode *> parent;
    map<int, int> depth;
    
    void dfs(TreeNode * node, int dpt){
        visited[node->val] = true;
        depth[node->val] = dpt;
        
        if(node->left != NULL){
            parent[node->left->val] = node;
            dfs(node->left, dpt + 1);
        }
        
        if(node->right != NULL){
            parent[node->right->val] = node;
            dfs(node->right, dpt + 1);
        }
    }
    
    TreeNode * lca(TreeNode * p, TreeNode * q){
        while(depth[p->val] != depth[q->val]){
            if(depth[p->val] > depth[q->val]){
                p = parent[p->val];
            }
            else{
                q = parent[q->val];
            }
        }
        
        while(p->val != q->val){
            p = parent[p->val];
            q = parent[q->val];
        }
        
        return p;
    }
    
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        TreeNode * answer;
        
        if(root != NULL){
            dfs(root, 0);
            answer = lca(p, q);
        }
        
        return answer;
    }
};