[LeetCode] BFS - 285. Inorder Successor in BST

문제

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:

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:

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, 104].
• -105 <= Node.val <= 105
• All Nodes will have unique values.

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코드

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @return {TreeNode}
 */

// p 다음으로 큰 숫자를 찾아라
var inorderSuccessor = function(root, p) {
    if (p.right) {
        // element is most left of the right sub-tree
        let subRoot = p.right;
        while (subRoot.left) {
            subRoot = subRoot.left;
        }
        return subRoot;
    } else {
        // element is last "left turned" node
        let lastLeft = null;
        let subRoot = root;
        while (subRoot && (subRoot.val != p.val)) {
            if (p.val < subRoot.val) {
                // turning left
                lastLeft = subRoot;
                subRoot = subRoot.left;
            } else {
                // turning right
                subRoot = subRoot.right;
            }
        }
        return lastLeft;
    }
};