[LeetCode285]Inorder Successor in BST

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.

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Hide Similar Problems (M) Binary Tree Inorder Traversal (M) Binary Search Tree Iterator

當然可以利用extra memory in order traverse整個bst再找node之後的那個值。 但其實我們可以O(n) w/o extra memory.

/**
 * 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* inorderSuccessor(TreeNode* root, TreeNode* p) {
        if (root == NULL || p == NULL) return NULL;

        TreeNode *suc = NULL;
        while (root != NULL) {
            if (root->val <= p->val) {
                root = root->right;
            } else {
                suc = root;
                root = root->left;
            }
        }
        return suc;
    }
};