99. 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?

解法1： In-order Traversal O(N)

BST的很重要的性质就是对他进行in order traversal的话，得到的是一个有序数组。
那么可以按照in order traversal,找到第一个node使得node.val >= node.successor.val和最后一个node使得node.predecessor.val >= node.val就可以了

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    TreeNode first = null;
    TreeNode second = null;
    TreeNode prev = null;
    public void recoverTree(TreeNode root) {
        if (root == null) return;
        inOrder(root);
        if (first == null || second == null) return;
        int temp = first.val;
        first.val = second.val;
        second.val = temp;
        return;
    }
    private void inOrder(TreeNode root) {
        if (root == null) return;
        inOrder(root.left);
        if (prev != null && prev.val >= root.val) {
            if (first == null) {
                first = prev;
            }
            if (first != null) {
                second = root;
            }
        }
        prev = root;
        inOrder(root.right);
    }
}