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
Solution 1: Binary Search
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:
- The value of the in-order successor node is greater than the value of node p.
- The in-order successor is the node with the smallest value among all nodes greater than p.
Therefore, for the current node root, if root.val > p.val, then root could be the in-order successor of p. We record root as ans and then search the left subtree, i.e., root = root.left. If root.val ≤ p.val, then root cannot be the in-order successor of p, and we search the right subtree, i.e., root = root.right.
The time complexity is O(h), where h is the height of the binary search tree. The space complexity is O(1).
PythonJavaC++GoTypeScriptJavaScript
# 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(code-box)
/**
* 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;
}
}(code-box)
/**
* 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;
}
};(code-box)
/**
* 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
}(code-box)
/**
* 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;
}(code-box)
/**
* 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;
};(code-box)
