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.
利用BST的性质来进行判断,如果两个子节点的最大值都小于root节点的值,那么他们的最低公共父节点一定在root节点的左侧,如果两个子节点的最小值都大于root节点值,那么他们的最低公共父节点一定在root节点的右侧。如果不符合以上2个情况,则root就是他们的最低公共父节点。
/** * 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 == nullptr || p == nullptr || q == nullptr) return root; if (max(p->val, q->val) < root->val) return lowestCommonAncestor(root->left, p, q); else if (min(p->val, q->val) > root->val) return lowestCommonAncestor(root->right, p, q); else return root; } }; // 29 ms