04.06. Successor

Description

Write an algorithm to find the "next" node (i.e., in-order successor) of a given node in a binary search tree.

Return null if there's no "next" node for the given node.

Example 1:

Input: root = [2,1,3], p = 1



  2

 / \

1   3



Output: 2

Example 2:

Input: root = [5,3,6,2,4,null,null,1], p = 6



      5

     / \

    3   6

   / \

  2   4

 /   

1



Output: null

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:

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 \(root\), 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 \(root.val \leq p.val\), 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 \(O(1)\).

Python3JavaC++GoTypeScriptJavaScriptSwift

# 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;
};

/* class TreeNode {
*    var val: Int
*    var left: TreeNode?
*    var right: TreeNode?
*
*    init(_ val: Int) {
*        self.val = val
*        self.left = nil
*        self.right = nil
*    }
* }
*/

class Solution {
    func inorderSuccessor(_ root: TreeNode?, _ p: TreeNode?) -> TreeNode? {
        var current = root
        var successor: TreeNode? = nil

        while let node = current {
            if node.val > p!.val {
                successor = node
                current = node.left
            } else {
                current = node.right
            }
        }
        return successor
    }
}

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