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.
思路:利用二叉搜索树的性质,可以判断出p和q与当前root的位置关系。若p和q的值都小于root,则它们的公共祖先一定在root的左子树;若p和q的值都大于root,则它们的公共祖先一定在root的右子树。
否则,q和q一定是一个在左子树一个在右子树,当前的root即是最小的公共祖先。
1 class Solution { 2 public: 3 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { 4 if (p->val < root->val && q->val < root->val) 5 return lowestCommonAncestor(root->left, p, q); 6 if (p->val > root->val && q->val > root->val) 7 return lowestCommonAncestor(root->right, p, q); 8 return root; 9 } 10 };