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:
二叉查找树,左子树的值比根值小,右子树的值比根值大。如果p q都比root小,说明在左子树上,如果pq的值都比root大说明在右子树上,如果一左一右,则root即为所求。
/** * 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) return lowestCommonAncestor(root -> left, p, q); if (p -> val > root -> val && q -> val > root -> val) return lowestCommonAncestor(root -> right, p, q); return root; } };
参考:
https://discuss.leetcode.com/topic/18614/easy-c-recursive-and-iterative-solutions