285. Inorder Successor in BST

JIAKAOBO

LeetCode

感谢赞助!

• ㊗️
• 大家
• offer
• 多多！

Problem

Given a binary search tree and a node in it, find the in-order successor of that node in the BST.

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.

Note:

If the given node has no in-order successor in the tree, return null. It’s guaranteed that the values of the tree are unique.

Code

class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        TreeNode res = null;

        while(root != null){
            if(p.val < root.val){
                res = root;
                root = root.left;
            } else {
                root = root.right;
            }
        }

        return res;
    }
}

class Solution {
    TreeNode pre;
    TreeNode res;
    
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        if(root == null) return null;
        if(root.left == null  && root.right == null) return null;

        helper(root, p);
        return res;
    }

    private void helper(TreeNode root, TreeNode target){
        if(root == null) return;

        helper(root.left, target);

        if(pre == target){
            res = root;
            // 不要忘记给pre赋值，不然走到下一步pre仍然是target
            pre = root;
            return;
        }

        pre = root;

        helper(root.right, target);
    }
}

class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        if(root == null) return null;

        if(p.val < root.val){
            // 结果一定在左边，但是当前的点可能是结果
            TreeNode temp = inorderSuccessor(root.left, p);
            // temp == null 表示没找到，所以就是当前点
            return (temp == null) ? root : temp;
        } else {
            return inorderSuccessor(root.right, p);
        }
    }
}

class Solution:
    def inorderSuccessor(self, root: 'TreeNode', p: 'TreeNode') -> 'TreeNode':
        res = None

        while root:
            if root.val > p.val:
                # 预留res为可能的结果
                res = root
                root = root.left
            else:
                root = root.right

        return res

class Solution:
    def inorderSuccessor(self, root: 'TreeNode', p: 'TreeNode') -> 'TreeNode':
        stack = []
        prev = False
        while root or len(stack):
            while root:
                stack.append(root)
                root = root.left

            root = stack.pop()
            if prev:
                return root

            if root.val == p.val:
                prev = True

            root = root.right

        return None