285. Inorder Successor in BST

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

Note: If the given node has no in-order successor in the tree, return null.

解法1： O(N) Space

最直观的方法，就是先做一遍in order, 存起来，然后找出来目标node前一个node即可。
C++

1

Java

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        List<TreeNode> list = new ArrayList<TreeNode>();
        if (root == null || p == null) {
            return null;
        }
        helper(root, list);
        for (int i = 0; i < list.size(); i++) {
            if (list.get(i) == p) {
                if (i == list.size() - 1) {
                    return null;
                } else {
                    return list.get(i + 1);
                }
            }
        }
        return null;
    }
    private void helper(TreeNode root, List<TreeNode> list) {
        if (root == null) {
            return;
        }
        helper(root.left, list);
        list.add(root);
        helper(root.right, list);
    }
}

解法2： O(1) Space

lang: java
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    TreeNode prev = null;
    TreeNode res = null;
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        if (root == null || p == null) {
            return null;
        }
        helper(root, p);
        return res;
    }
    private void helper(TreeNode root, TreeNode p) {
        if (root == null) {
            return;
        }
        helper(root.left, p);
        if (prev == p) {
            res = root;
            prev = root;
            return;
        }
        prev = root;
        helper(root.right, p);
    }
}

解法3: Iterative

lang: java
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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) {
        if (root == null) return root;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode prev = null;
        while (!stack.isEmpty() || root != null) {
            if (root != null) {
                stack.push(root);
                root = root.left;
            } else {
                root = stack.pop();
                if (prev == p) return root;            
                prev = root;
                root = root.right;
            }
        }
        return null;
    }
}