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1.分别和当前节点比较(最开始是根节点)如果p->val和q->val一个大于当前节点值,一个小于当前节点值,则表示当前节点是他们的最低公共祖先,如果两个的大于当前节点的值,则递归调用右子树,如果都小于,则递归调用左子树。
/** * 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(p->val > root->val&&q->val > root->val) { root = lowestCommonAncestor(root->right,p,q); } if(p->val < root->val&&q->val < root->val) { root = lowestCommonAncestor(root->left,p,q); } return root; } };