Skip to content

285. Inorder Successor in BST πŸ”’

DifficultyMedium

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:

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.

Solutions

Thinking

The inorder successor is the smallest key greater than \(p\). On a BST we walk as in binary search: if the current key is larger it is a candidate and we go left; otherwise we go right.

The last candidate on that path is the successor, in time proportional to the height.

The in-order traversal of a binary search tree is an ascending sequence, so we can use the binary search method.

The in-order successor node of a binary search tree node \(p\) satisfies:

  1. The value of the in-order successor node is greater than the value of node \(p\).
  2. The in-order successor is the node with the smallest value among all nodes greater than \(p\).

Therefore, for the current node \(\textit{root}\), if \(\textit{root}.\textit{val} > p.\textit{val}\), then \(\textit{root}\) could be the in-order successor of \(p\). We record \(\textit{root}\) as \(\textit{ans}\) and then search the left subtree, i.e., \(\textit{root} = \textit{root}.\textit{left}\). If \(\textit{root}.\textit{val} \leq p.\textit{val}\), then \(\textit{root}\) cannot be the in-order successor of \(p\), and we search the right subtree, i.e., \(\textit{root} = \textit{root}.\textit{right}\).

The time complexity is \(O(h)\), and the space complexity is \(O(1)\), where \(h\) is the height of the binary search tree.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution:
    def inorderSuccessor(self, root: TreeNode, p: TreeNode) -> Optional[TreeNode]:
        ans = None
        while root:
            if root.val > p.val:
                ans = root
                root = root.left
            else:
                root = root.right
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        TreeNode ans = null;
        while (root != null) {
            if (root.val > p.val) {
                ans = root;
                root = root.left;
            } else {
                root = root.right;
            }
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
/**
 * 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) {
        TreeNode* ans = nullptr;
        while (root) {
            if (root->val > p->val) {
                ans = root;
                root = root->left;
            } else {
                root = root->right;
            }
        }
        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func inorderSuccessor(root *TreeNode, p *TreeNode) (ans *TreeNode) {
    for root != nil {
        if root.Val > p.Val {
            ans = root
            root = root.Left
        } else {
            root = root.Right
        }
    }
    return
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function inorderSuccessor(root: TreeNode | null, p: TreeNode | null): TreeNode | null {
    let ans: TreeNode | null = null;
    while (root) {
        if (root.val > p.val) {
            ans = root;
            root = root.left;
        } else {
            root = root.right;
        }
    }
    return ans;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
/**
 * 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}
 */
var inorderSuccessor = function (root, p) {
    let ans = null;
    while (root) {
        if (root.val > p.val) {
            ans = root;
            root = root.left;
        } else {
            root = root.right;
        }
    }
    return ans;
};

Comments