[LeetCode] Lowest Common Ancestor of a Binary Search Tree

时间：2022-08-26 10:05:35

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.

 

经典问题！

方法一：找到两个节点的路径，然后根据路径找LCA。

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     void getPath(TreeNode *root, TreeNode *p, vector<TreeNode*> &path) {
13         TreeNode *tmp = root;
14         while (tmp != p) {
15             path.push_back(tmp);
16             if (tmp->val > p->val) tmp = tmp->left;
17             else tmp = tmp->right;
18         }
19         path.push_back(p);
20     }
21     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
22         vector<TreeNode*> path1, path2;
23         getPath(root, p, path1);
24         getPath(root, q, path2);
25         TreeNode *res = root;
26         int idx = 0;
27         while (idx < path1.size() && idx < path2.size()) {
28             if (path1[idx] != path2[idx]) break;
29             else res = path1[idx++];
30         }
31         return res;
32     }
33 };

 

方法二：根据BST的性质，两个节点a，b的公共袓先c一定满足a <= c <= b 或者 a >= c >= b。

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
13         TreeNode *cur = root;
14         while (cur != NULL) {
15             if (cur->val > p->val && cur->val > q->val) cur = cur->left;
16             else if (cur->val < p->val && cur->val < q->val) cur = cur->right;
17             else return cur;
18         }
19         return cur;
20     }
21 };

 

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