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

题目描述

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.

解题思路 —— 递归求解

设 x 为 p，q较大值节点，y 为 p，q较小值节点。则当遍历到 r 节点时，它们的位置关系有以下几种可能性：

• “x 节点 是 r 节点且 y 节点是 x 节点的左子树的节点”或者“y节点 是 r 节点且 x 节点是 y 节点的右子树的节点”
  这种情况下，最近的公共祖先节点就是 x 节点或 y 节点。
  

• y 节点和 x 节点分别是 r 节点的左/右子树的节点
  这种情况下，最近的公共祖先节点就是 r 节点。
  

• x 节点和 y 节点都是 r 节点的左子树的节点
  这种情况下，最近的公共祖先节点在 r 节点的左子树中。
  

• x 节点和 y 节点都是 r 节点的右子树的节点
  这种情况下，最近的公共祖先节点在 r 节点的右子树中。
  

AC 代码

/**
* 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 == nullptr || p == nullptr || q == nullptr)
       {
           return nullptr;
       }
       
       int maxVal = max(p->val, q->val);
       int minVal = min(p->val, q->val);
       
       //  “`x` 节点 是 `r` 节点且 `y` 节点是 `x` 节点的左子树的节点”
       // 或者“`y`节点 是 `r` 节点且 `x` 节点是 `y` 节点的右子树的节点”
       if(root->val == p->val || root->val == q->val)
       {
           return root;
       }
       // `y` 节点和 `x` 节点分别是 `r` 节点的左/右子树的节点
       else if(root->val < maxVal && root->val > minVal)
       {
           return root;
       }
       // `x` 节点和 `y` 节点都是 `r` 节点的左子树的节点
       else if(root->val > maxVal)
       {
           return lowestCommonAncestor(root->left, p, q);
       }
       //  `x` 节点和 `y` 节点都是 `r` 节点的右子树的节点
       else //if(root->val < minVal)
       {
           return lowestCommonAncestor(root->right, p, q);
       }
   }
};