LeetCode刷题笔录Recover Binary Search Tree

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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.

和Valid Binary Search Tree那题比较像，也是要做一个in order traversal，过程中比较当前node的值和之前一个node的值得大小，看看有没有反序的情况。

有两种情况：1是两个相邻的元素被交换了，比如12345678 -> 12354678，反序只会发生一次，因此记录5，4两个node就对了

2是不相邻元素被交换了，比如12345678 -> 12745638,反序会发生两次，一次在74一次在63.可以看出需要交换的元素是7和3，即第一次反序的前一个元素和第二次反序的后一个元素。

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public void recoverTree(TreeNode root) {        ArrayList<TreeNode> pre = new ArrayList<TreeNode>();        pre.add(null);        ArrayList<TreeNode> nodes = new ArrayList<TreeNode>();        inorder(root, pre, nodes);                int temp = nodes.get(0).val;        nodes.get(0).val = nodes.get(1).val;        nodes.get(1).val = temp;    }        public void inorder(TreeNode root, ArrayList<TreeNode> pre, ArrayList<TreeNode> nodes){        if(root == null)            return;                inorder(root.left, pre, nodes);                if(pre.get(0) != null && root.val < pre.get(0).val){            if(nodes.size() == 0){                nodes.add(pre.get(0));                nodes.add(root);            }            else{                nodes.set(1, root);            }        }        pre.set(0, root);        inorder(root.right, pre, nodes);            }}

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