力扣0099——恢复二叉搜索树

恢复二叉搜索树

难度：中等

题目描述

给你二叉搜索树的根节点 root ，该树中的 恰好 两个节点的值被错误地交换。请在不改变其结构的情况下，恢复这棵树 。

示例1

输入： root = [1,3,null,null,2]
输出：[3,1,null,null,2]

示例2

输入： root = [3,1,4,null,null,2]
输出：[2,1,4,null,null,3]

题解

因为二叉搜索树的性质可得，将其中序遍历存储到列表中，数值为单调递增，由此可以得到以下步骤

• 遍历列表，找到递增中断点
• 再次遍历列表，找到中断点应该在的位置
• 将两个数值进行交换

完成之后即为所求

想法代码

public class TreeNode
{
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int val = 0, TreeNode left = null, TreeNode right = null)
    {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}
class Solution
{
    IList<TreeNode> travelList = new List<TreeNode>();
    public static void Main(String[] args)
    {
        TreeNode root = new TreeNode(3)
        {
            left = new TreeNode(1),
            right = new TreeNode(4)
            {
                left = new TreeNode(2)
            }

        };
        Solution solution = new Solution();
        solution.RecoverTree(root);
        foreach (var a in solution.travelList)
        {
            Console.Write(a.val + " ");
        }
    }

    public void RecoverTree(TreeNode root)
    {
        Travel(root);
        int index1 = 1;
        int index2 = 0;
        while (index1 < travelList.Count)
        {
            if (travelList[index1].val > travelList[index1 - 1].val)
            {
                index1++;
            }
            else
            {
                break;
            }
        }

        while (index2 < travelList.Count)
        {
            if (travelList[index2].val > travelList[index1 - 1].val)
            {
                break;
            }

            index2++;
        }
        TreeNode treeNode1 = travelList[index1 - 1];
        TreeNode treeNode2 = travelList[index2 - 1];
        int val1 = treeNode1.val;
        int val2 = treeNode2.val;
        treeNode1.val = val2;
        treeNode2.val = val1;
    }

    public void Travel(TreeNode root)
    {
        if (root == null)
        {
            return;
        }

        Travel(root.left);
        travelList.Add(root);
        Travel(root.right);
    }
}
avel(root.left);
        travelList.Add(root);
        Travel(root.right);
    }
}