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 nodes2and8is6.
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 nodes2and4is2, 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.
这道题让我们求二叉搜索树的最小共同父节点, LeetCode中关于BST的题有 Kth Smallest Element in a BST。这道题我们可以用递归来求解，我们首先来看题目中给的例子，由于二叉搜索树的特点是左<根<右，所以根节点的值一直都是中间值，大于左子树的所有节点值，小于右子树的所有节点值，那么我们可以做如下的判断，如果根节点的值大于p和q之间的较大值，说明p和q都在左子树中，那么此时我们就进入根节点的左子节点继续递归，如果根节点小于p和q之间的较小值，说明p和q都在右子树中，那么此时我们就进入根节点的右子节点继续递归，如果都不是，则说明当前根节点就是最小共同父节点，直接返回即可，参见代码如下：
解法一：
class Solution { public: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { if (!root) return NULL; if (root->val > max(p->val, q->val)) return lowestCommonAncestor(root->left, p, q); else if (root->val < min(p->val, q->val)) return lowestCommonAncestor(root->right, p, q); else return root; } };
当然，此题也有非递归的写法，用个 while 循环来代替递归调用即可，然后不停的更新当前的根节点，也能实现同样的效果，代码如下：
解法二：
class Solution { public: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { while (true) { if (root->val > max(p->val, q->val)) root = root->left; else if (root->val < min(p->val, q->val)) root = root->right; else break; } return root; } };
Github 同步地址：
https://github.com/grandyang/leetcode/issues/235
类似题目：
Lowest Common Ancestor of a Binary Tree
参考资料：
https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
LeetCode All in One 题目讲解汇总(持续更新中...)