Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the definition of LCA on Wikipedia: "The lowest common ancestor is defined between two nodes p
and q
as the lowest node in T
that has both p
and q
as descendants (where we allow a node to be a descendant of itself)."
Example 1:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.
Example 2:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
Example 3:
Input: root = [2,1], p = 2, q = 1
Output: 2
Constraints:
- The number of nodes in the tree is in the range
[2, 105]
. -109 <= Node.val <= 109
- All
Node.val
are unique. p != q
p
andq
will exist in the BST.
递归法:
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(root->val>p->val && root->val>q->val){
return lowestCommonAncestor(root->left,p,q);
}
else if(root->val<p->val && root->val<q->val){
return lowestCommonAncestor(root->right,p,q);
}
else return root;
}
};
注意:
第一步,如果root->val比pq都要大就递归左子树,反之递归右子树
第二步,如果不是上面两种(root就是最近公共祖先)就返回root
迭代:
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
while(root){
if(root->val>p->val && root->val>q->val)root=root->left;
else if(root->val<p->val && root->val<q->val)root=root->right;
else return root;
}
return NULL;
}
};
注意:
不要忘记最后一行的return NULL,又是被这么简短的迭代感动哭的一天