285. Inorder Successor in BST

题目：

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.

链接： http://leetcode.com/problems/inorder-successor-in-bst/

题解：

一开始的想法就是用inorder traversal，设置一个boolean变量，当找到root.val = p.val的时候返回下一个节点，遍历完毕以后返回null。

Time Complexity - O(n)， Space Complexity - O(n)

/**
 * 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) {
        if(root == null || p == null) {
            return null;
        }
        boolean foundNodeP = false;
        Stack<TreeNode> stack = new Stack<>();
        while(root != null || !stack.isEmpty()) {
            if(root != null) {
                stack.push(root);
                root = root.left;
            } else {
                root = stack.pop();
                if(foundNodeP) {
                    return root;
                }
                if(root.val == p.val) {
                    foundNodeP = true;
                }
                root = root.right;
            }
        }
        
        return null;
    }
}

 

看了Discuss之后发现有很多简洁的写法，而且不用遍历全部元素。利用BST的性质，比较root.val和p.val，然后在左子树或者右子树中查找。

Time Complexity - O(h)， Space Complexity - O(1)

/**
 * 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) {
        if(root == null || p == null) {
            return null;
        }
        TreeNode successor = null;
        while(root != null) {
            if(p.val < root.val) {
                successor = root;
                root = root.left;
            } else {
                root = root.right;
            }
        }
        
        return successor;
    }
}

 

二刷:

方法和一刷一样。我们先建立一个空的successor，再取得一个root节点的reference。每次当node.val > p.val的时候，我们记录下当前的node节点，然后往左子树查找。否则向右子树查找。 向右子树查找的过程中不需要更新successor。

Java:

Time Complexity - O(h)， Space Complexity - O(1)

/**
 * 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) {
        TreeNode node = root, successor = null;
        while (node != null) {
            if (node.val > p.val) {
                successor = node;
                node = node.left;
            } else {
                node = node.right;
            }
        }
        return successor;
    }
}

 

 

 

Reference:

https://leetcode.com/discuss/69200/for-those-who-is-not-so-clear-about-inorder-successors

https://leetcode.com/discuss/61105/java-python-solution-o-h-time-and-o-1-space-iterative

https://leetcode.com/discuss/59728/10-and-4-lines-o-h-java-c

https://leetcode.com/discuss/59787/share-my-java-recursive-solution