235. Lowest Common Ancestor of a Binary Search Tree
Total Accepted: 77021 Total Submissions: 205265 Difficulty: Easy
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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”
_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5
For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
思路：
二叉排序树(Binary Sort Tree)又称二叉查找(搜索)树(Binary Search Tree)。其定义为：二叉排序树或者是空树，或者是满足如下性质的二叉树：
①若它的左子树非空，则左子树上所有结点的值均小于根结点的值；
②若它的右子树非空，则右子树上所有结点的值均大于根结点的值；
③左、右子树本身又各是一棵二叉排序树。
代码：
迭代：
1 /**
2 * Definition for a binary tree node.
3 * struct TreeNode {
4 * int val;
5 * TreeNode *left;
6 * TreeNode *right;
7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8 * };
9 */
10 class Solution {
11 public:
12 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
13 while(true){
14 if(root->val>p->val&&root->val>q->val){
15 root=root->left;
16 continue;
17 }
18 if(root->val<p->val&&root->val<q->val){
19 root=root->right;
20 continue;
21 }
22 return root;
23 }
24 }
25 };
递归：
1 class Solution {
2 public:
3 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
4 if(root->val>p->val&&root->val>q->val) return lowestCommonAncestor(root->left,p,q);
5 if(root->val<p->val&&root->val<q->val) return lowestCommonAncestor(root->right,p,q);
6 return root;
7 }
8 };
还有另外1题，和这题相似：Leetcode 236. Lowest Common Ancestor of a Binary Tree
来源：https://www.cnblogs.com/Deribs4/p/5626215.html