题目来源
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.
求最低的公共父节点。没想到应该怎么做。
发现自己有点傻，那个这是个二叉搜索树啊，这树的特性你得利用起来啊！
就是假如p、q在同一个子树下的话，那么它们要么都比root节点大，要么都比root节点小。
代码如下：
/**
* 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 == NULL)
return NULL;
if (p->val < root->val && q->val < root->val)
return lowestCommonAncestor(root->left, p, q);
if (p->val > root->val && q->val > root->val)
return lowestCommonAncestor(root->right, p, q);
return root;
}
};
