Given a binary search tree and a node in it, find the in-order successor of that node in the BST.
The successor of a node p is the node with the smallest key greater than p.val.
You will have direct access to the node but not to the root of the tree. Each node will have a reference to its parent node.
Example 1:
Input: root = {"$id":"1","left":{"$id":"2","left":null,"parent":{"$ref":"1"},"right":null,"val":1},"parent":null,"right":{"$id":"3","left":null,"parent":{"$ref":"1"},"right":null,"val":3},"val":2} p = 1 Output: 2 Explanation: 1's in-order successor node is 2. Note that both p and the return value is of Node type. Example 2:
Input: root = {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":{"$id":"4","left":null,"parent":{"$ref":"3"},"right":null,"val":1},"parent":{"$ref":"2"},"right":null,"val":2},"parent":{"$ref":"1"},"right":{"$id":"5","left":null,"parent":{"$ref":"2"},"right":null,"val":4},"val":3},"parent":null,"right":{"$id":"6","left":null,"parent":{"$ref":"1"},"right":null,"val":6},"val":5} p = 6 Output: null Explanation: There is no in-order successor of the current node, so the answer is null. Example 3:
Input: root = {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":{"$id":"4","left":null,"parent":{"$ref":"3"},"right":null,"val":2},"parent":{"$ref":"2"},"right":{"$id":"5","left":null,"parent":{"$ref":"3"},"right":null,"val":4},"val":3},"parent":{"$ref":"1"},"right":{"$id":"6","left":null,"parent":{"$ref":"2"},"right":{"$id":"7","left":{"$id":"8","left":null,"parent":{"$ref":"7"},"right":null,"val":9},"parent":{"$ref":"6"},"right":null,"val":13},"val":7},"val":6},"parent":null,"right":{"$id":"9","left":{"$id":"10","left":null,"parent":{"$ref":"9"},"right":null,"val":17},"parent":{"$ref":"1"},"right":{"$id":"11","left":null,"parent":{"$ref":"9"},"right":null,"val":20},"val":18},"val":15} p = 15 Output: 17 Example 4:
Input: root = {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":{"$id":"4","left":null,"parent":{"$ref":"3"},"right":null,"val":2},"parent":{"$ref":"2"},"right":{"$id":"5","left":null,"parent":{"$ref":"3"},"right":null,"val":4},"val":3},"parent":{"$ref":"1"},"right":{"$id":"6","left":null,"parent":{"$ref":"2"},"right":{"$id":"7","left":{"$id":"8","left":null,"parent":{"$ref":"7"},"right":null,"val":9},"parent":{"$ref":"6"},"right":null,"val":13},"val":7},"val":6},"parent":null,"right":{"$id":"9","left":{"$id":"10","left":null,"parent":{"$ref":"9"},"right":null,"val":17},"parent":{"$ref":"1"},"right":{"$id":"11","left":null,"parent":{"$ref":"9"},"right":null,"val":20},"val":18},"val":15} p = 13 Output: 15
Note:
If the given node has no in-order successor in the tree, return null. It's guaranteed that the values of the tree are unique. Remember that we are using the Node type instead of TreeNode type so their string representation are different.
// 732ms, 59%
/*
// Definition for a Node.
class Node {
public:
int val;
Node* left;
Node* right;
Node* parent;
};
*/
class Solution {
public:
Node* inorderSuccessor(Node* node) {
if (!node) return nullptr;
// has right child
if (node->right) {
node = node->right;
while (node->left)
node = node->left;
return node;
}
/* no need to check here
// is root of the tree
if (!node->parent)
return nullptr;
// is left child
if (node == node->parent->left)
return node->parent;
*/
// is right child
while (node->parent && node == node->parent->right)
node = node->parent;
return node->parent;
}
};