235. Lowest Common Ancestor of a Binary Search Tree

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.

解题思路：

因为是二叉搜索树，有左子树的所有值都小于父节点的值，右子树的所有值都大于父节点，因此递归判断两个节点在左子树还是右子树，若在两边，则父节点就是其最小公共祖先，否则递归搜索同在一遍的子树。

#include<windows.h> using namespace std; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; class Solution { public: // 40ms-42ms TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { if(root==NULL)return root; if(p==NULL)return q; if(q==NULL)return p; int val=root->val; if((val-p->val)*(val-q->val)<=0)return root; if(val>p->val)return lowestCommonAncestor(root->left,p,q); return lowestCommonAncestor(root->right,p,q); } };