【LeetCode】: 235. Lowest Common Ancestor of a Binary Search Tree

tags: LeetCode C++ Binary Search Tree

Description

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

取兩個 node 的最小祖先。

https://en.wikipedia.org/wiki/Lowest_common_ancestor

Solution

因為是 BST 所以可以比大小，判斷如果兩個 node 分在判斷 node 的兩側，則 LCA 就一定是該 node，若落在兩邊，則更新當前 node 到那一側繼續判斷。

如果有 node 等於當前 node 的話則回傳當前 node。
























/**
 * 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) {
        while(root!=nullptr){
            if(root->val == p->val || root->val == q->val) return root;
            if(root->val > p->val && root->val < q->val) return root;
            if(root->val < p->val && root->val > q->val) return root;
            if(root->val < p->val && root->val < q->val) root = root->right;
            if(root->val > p->val && root->val > q->val) root = root->left;
        }
        return root;
    }
};

也可以利用 Recursion 的方式遞迴查找。



















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