Leetcode: Recover Binary Search Tree

LuckySeven7777

于 2015-01-28 04:10:16 发布

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本文链接：https://blog.youkuaiyun.com/LuckySeven7777/article/details/43222669

本文详细介绍了如何使用中序遍历来修复二叉搜索树中两个元素位置错误的问题，提供了常数空间解决方案，并解释了序列化和反序列化二叉树的基本概念。

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Problem:

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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

Solution:

Using Inorder Traverse.

Code

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    TreeNode p;
    TreeNode q;
    TreeNode tmp = new TreeNode(Integer.MIN_VALUE);
    public void recoverTree(TreeNode root) {
        if (root == null)  return;
        recover(root);
        if (p != null && q != null) {
            int t = p.val;
            p.val = q.val;
            q.val = t;
        }
        
    }
    public void recover(TreeNode root) {
        if (root == null)  return;
        recover(root.left);
        if (tmp != null && root.val < tmp.val) {
            if(p == null) {
                p = tmp;
                q = root;
            } else {
                q = root;
            }
        }
        tmp = root;
        recover(root.right);
    }
}