24
07/2015
[LeetCode] Lowest Common Ancestor of a Binary Search Tree
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：
本题题意是找到二分查找树的两个节点的最低公共祖先。根据二分查找树的特点，两个节点的最低公共祖先一定是大于等于较小节点，并且小于等于较大节点。倘若某个节点N的值都比给定的两个节点的值大，则公共节点一定在N的左孩子节点中。倘若某个节点N的值都比给定的两个节点的值小，则公共节点一定在N的右孩子节点中。
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) {
int minNum = min(p->val, q->val);
int maxNum = max(p->val, q->val);
while(root!=NULL){
if(minNum <= root->val && root->val <= maxNum)
{
return root;
}else if(root->val < minNum){
root = root->right;
}else{
root = root->left;
}
}
return NULL;
}
};
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