15算法课程 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.

solution:

终止条件

1）根节点直接为NULL；

2）节点一个在左，一个在右，或到最后其中一个根本就是根节点本身。

递归过程

1）都在左子树中，需要在左子树中继续搜索公共祖先

2）都在右子树中，需要在右子树中继续搜索公共祖先

code:

class Solution {
public:
// 给定一棵二分搜索树，求两个节点的最近公共祖先
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
// 递归终止条件
if (root == NULL || p == NULL || q == NULL) return NULL;

// 找到公共祖先的情况
if (p->val <= root->val && q->val >= root->val || p->val >= root->val && q->val <= root->val)
return root;

// 左子树中递归
if (p->val < root->val && q->val < root->val)
return lowestCommonAncestor(root->left, p, q);

// 右子树中递归
if (p->val>root->val && q->val>root->val)
return lowestCommonAncestor(root->right, p, q);
}
};