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#
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| # 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
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| /**
* 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;
}
}
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| /**
* 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;
}
};
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| /**
* 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
}
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| /**
* 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;
}
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| /**
* 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;
};
|
