[LeetCode] 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.
分析:二叉搜索树是排序过的,位于左子树的节点都比父节点小,而位于右子树的节点都比父节点大,我们只需要从树的根节点开始和两个输入的节点比较。如果当前节点的值比两个节点的值都大,那么最低公共祖先一定是在当前节点的左子树中,于是下一步遍历当前节点的左子树。如果当前节点的值比两个节点的值都小,那么最低公共祖先一定在当前节点的右子树中,于是下一步遍历当前节点的右子树。这样在树中从上到下找到的第一个在两个输入节点的值之间的节点,就是最低的公共祖先。
/** * 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 (root->val < p->val && root->val < q->val) { return lowestCommonAncestor(root->right, p, q); } else if (root->val > p->val && root->val > q->val) { return lowestCommonAncestor(root->left, p, q); } else { return root; } } };