Leetcode之Lowest Common Ancestor of a Binary Search Tree

題目：

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).”

Given binary search tree:  root = [6,2,8,0,4,7,9,null,null,3,5]

 

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.

 

Note:

• All of the nodes' values will be unique.
• p and q are different and both values will exist in the BST.

代碼：

方法一——直接利用二叉搜索樹的性質：

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

方法二——兩邊分開找：

 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
       if (!root || p == root || q == root) return root;
       TreeNode *left = lowestCommonAncestor(root->left, p, q);
       TreeNode *right = lowestCommonAncestor(root->right, p , q);
       if (left && right) return root;
       return left ? left : right;
    }