Question

285. Inorder Successor in BST

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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

# 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

/**
 * 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;
  }
}

/**
 * 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;
  }
};

/**
 * 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
}

/**
 * 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;
}

/**
 * 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;
};