0897. Increasing Order Search Tree (M)

Increasing Order Search Tree (M)

题目

Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.

Example 1:

Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

Example 2:

Input: root = [5,1,7]
Output: [1,null,5,null,7] 

Constraints:

• The number of nodes in the given tree will be in the range [1, 100].
• 0 <= Node.val <= 1000

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题意

将一个BST按照顺序变成一直链。

思路

简单的可以用中序遍历实现，或者用类似Morris遍历的方法也可以。

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代码实现

Java

中序遍历

class Solution {
    public TreeNode increasingBST(TreeNode root) {
        List<TreeNode> list = new ArrayList<>();
        inorder(root, list);

        TreeNode dummy = new TreeNode();
        TreeNode p = dummy;
        for (TreeNode node : list) {
            p.right = node;
            p = node;
        }
        return dummy.right;
    }

    private void inorder(TreeNode root, List<TreeNode> list) {
        if (root == null) {
            return;
        }

        inorder(root.left, list);
        list.add(root);
        root.left = null;
        inorder(root.right, list);
    }
}

Morris

class Solution {
    public TreeNode increasingBST(TreeNode root) {
        TreeNode newRoot = root;

        while (newRoot != null && newRoot.left != null) {
            newRoot = newRoot.left;
        }

        while (root != null) {
            if (root.left != null) {
                TreeNode rightMost = root.left;
                while (rightMost.right != null) {
                    rightMost = rightMost.right;
                }
                rightMost.right = root;
                root = root.left;
                rightMost.right.left = null;
            } else if (root.right != null && root.right.left != null) {
                root.right = increasingBST(root.right);
                root = root.right;
            } else {
                root = root.right;
            }
        }

        return newRoot;
    }
}