[LeetCode] Inorder Successor in BST

Problem Description:

Given a binary search tree and a node in it, find the in-order successor of that node in the BST.

Note: If the given node has no in-order successor in the tree, return null.

---

There are just two cases:

1. The easier one: p has right subtree, then its successor is just the leftmost child of its right subtree;
2. The harder one: p has no right subtree, then a traversal is needed to find its successor.

Traversal: we start from the root, each time we see a node with val larger than p -> val, we know this node may be p's successor. So we record it in suc. Then we try to move to the next level of the tree: if p -> val > root -> val, which means p is in the right subtree, then its successor is also in the right subtree, so we update root = root -> right; if p -> val < root -> val, we update root = root -> left similarly; once we find p -> val == root -> val, we know we've reached at p and the current sucis just its successor.

The code is as follows. You may try some examples to see how it works :-)

class Solution {
public:
    TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
        if (p -> right) return leftMost(p -> right);
        TreeNode* suc = NULL;
        while (root) {
            if (p -> val < root -> val) {
                suc = root;
                root = root -> left;
            }
            else if (p -> val > root -> val)
                root = root -> right; 
            else break;
        }
        return suc;
    }
private:
    TreeNode* leftMost(TreeNode* node) {
        while (node -> left) node = node -> left;
        return node;
    }
};