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剑指 Offer 68 - I. 二叉搜索树的最近公共祖先

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Symbols count in article: 1.3k Reading time ≈ 1 mins.

剑指 Offer 68 - I. 二叉搜索树的最近公共祖先

参考236. Lowest Common Ancestor of a Binary Tree

但因为这题是BST，所以有别的方法

以下情况:

  1. 当p，q都在root左侧，那么p，q都在root左子树。
  2. 当p，q都在root右侧，那么p，q都在root右子树。
  3. 当第一次产生没在同一个子树时，这时候的root就是解
    1. p或q为root，剩下的在一边的树
    2. p，q在root两侧
迭代
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/**
 * 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)
        {
            if(root->val > p->val && root->val > q->val)
                root = root->left;
            else if(root->val < p->val && root->val < q->val)
                root = root->right;
            else
                return root;
        }
        return nullptr;
    }
};
递归
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/**
 * 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->left, p, q);
            else if(root->val < p->val && root->val < q->val)
                return lowestCommonAncestor(root->right, p, q);
            return root;
    }
};