285. Inorder Successor in BST

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285. Inorder Successor in BST

Description

Given the root of a binary search tree and a node p in it, return the in-order successor of that node in the BST. If the given node has no in-order successor in the tree, return null.

The successor of a node p is the node with the smallest key greater than p.val.

Example 1:

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Input: root = [2,1,3], p = 1
Output: 2
Explanation: 1's in-order successor node is 2. Note that both p and the return value is of TreeNode type.

Example 2:

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Input: root = [5,3,6,2,4,null,null,1], p = 6
Output: null
Explanation: There is no in-order successor of the current node, so the answer is `null`.

Constraints:

  • The number of nodes in the tree is in the range [1, 10^4].
  • -10^5 <= Node.val <= 10^5
  • All Nodes will have unique values.

Hints/Notes

  • binary search tree

Solution

Language: C++

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/**
 * 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* node = nullptr;
    int found = false;

    TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
        traverse(root, p);
        return node;
    }

    void traverse(TreeNode* root, TreeNode* p) {
        if (!root || node) {
            return;
        }
        traverse(root->left, p);

        if (found && !node) {
            node = root;
        }

        if (root == p) {
            found = true;
        }
        traverse(root->right, p);
    }
};