[leetcode] Recover Binary Search Tree

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

https://oj.leetcode.com/problems/recover-binary-search-tree/

 

思路1：中序遍历生成序列，然后对其中错序的进行调整。 需要额外O(n)的空间。

思路2：中序遍历二叉树，记录当前指针cur的前一个节点pre，如果pre.val大于cur.val，表示有错序，多数情况错序有两次；如果有一次错序，说明就是相邻节点需要被交换。

 

public class Solution {

    private TreeNode pre;
    private TreeNode wrong1;
    private TreeNode wrong2;

    public void recoverTree(TreeNode root) {
        preOrder(root);
        int tmp = wrong1.val;
        wrong1.val = wrong2.val;
        wrong2.val = tmp;

    }

    private void preOrder(TreeNode root) {
        if (root == null)
            return;
        preOrder(root.left);

        if (pre != null && pre.val > root.val) {
            if (wrong1 == null) {
                wrong1 = pre;
                wrong2 = root;
            } else {
                wrong2 = root;
            }

        }

        pre = root;

        preOrder(root.right);

    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(10);
        root.left = new TreeNode(5);
        root.right = new TreeNode(15);
        root.left.right = new TreeNode(13);
        root.right.left = new TreeNode(7);

        new Solution().recoverTree(root);

    }

}

 

第二遍记录：注意记录pre的用法。

 

第三遍记录：

　　举一个例子，中序遍历后正常应该是递增的，随意交换两个元素，会发现两种情况

　　交换相邻元素后，序列只有一处地方递减。

　　交换非相邻元素，序列有两处地方递减。

public class Solution {

    private TreeNode one;
    private TreeNode two;
    private TreeNode pre;

    public void recoverTree(TreeNode root) {
        inorder(root);
        int tmp = one.val;
        one.val = two.val;
        two.val = tmp;

    }

    private void inorder(TreeNode root) {
        if (root != null) {
            inorder(root.left);

            if (pre != null && pre.val > root.val) {
                if (one == null) {
                    one = pre;
                    two = root;
                } else {
                    two = root;
                }
            }

            pre = root;
            inorder(root.right);
        }
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(10);
        root.left = new TreeNode(5);
        root.right = new TreeNode(15);
        root.left.right = new TreeNode(13);
        root.right.left = new TreeNode(7);

        new Solution().recoverTree(root);

    }

}

 

 

参考：

http://blog.csdn.net/worldwindjp/article/details/21694179