LeetCode 0235 Lowest Common Ancestor of a Binary Search Tree【二叉搜索树，LCA】

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.

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题意

二叉搜索树的最近公共祖先

思路1

• 若p和q两个结点在当前根节点两侧，则当前根结点为所求
• 否则在两边子树中分别求

代码1

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