static void Main(string[] args)
{
/*
* Test case 1:
*/
TreeNode n1 = new TreeNode(0);
n1.left = new TreeNode(1);
//recoverTree(n1);
TreeNode n2 = new TreeNode(2);
n2.right = new TreeNode(1);
recoverTree(n2);
TreeNode n3 = new TreeNode(4);
n3.left = new TreeNode(6);
n3.left.left = new TreeNode(1);
n3.left.right = new TreeNode(3);
n3.right = new TreeNode(2);
n3.right.left = new TreeNode(5);
n3.right.right = new TreeNode(7);
//recoverTree(n3);
}
private static TreeNode v2 = null; // the second violation node
#endregion Fields
#region Methods
public static void recoverTree(TreeNode root)
{
// traverse and get two elements
traverse(root);
// swap
int temp = v1.val;
v1.val = v2.val;
v2.val = temp;
}
/*
* julia's comment:
* the violation in inorder traversal output is not written in most simple form.
* Rewrite the second version traverse_B
*/
private static void traverse(TreeNode t)
{
if (t == null)
{
return;
}
traverse(t.left);
bool inIncreasingOrder = t.val > theLast.val;
if (!inIncreasingOrder) // not in increasing order
{
if (v1 == null)
{
v1 = t;
v2 = theLast; // this is the one - violation one.
}
else
{
v1 = v2;
v2 = t;
}
}
theLast = t;
traverse(t.right);
}
/*
*
* Base test case:
* tree only have two nodes, switched;
* case A:
* 2 1
* / -> /
* 1 2
* case B:
* 1 2
* \ -> \
* 2 1
*
violation facts:
* 1. first node, it is the one with bigger value, violated;
* but second node, it is the one with smaller value, violated.
* 2. sometimes, only one violation catch; like base case A, switch two nodes, only two nodes
* in the tree. Tree [2,#,1]
* 3. sometimes, two violationes catch.
* Tree [2, 3, 1], inorder traversal result is 3, 2, 1, and then,
* first one of viloations is 3, 2, 3>2, the violation one is 3, bigger value;
* second one of viloations is 2, 1, 2>1, the violation one is 1, smaller value;
* so, 3 and 1 should be swapped back as 1 (2) 3
*/
private static void traverse_B(TreeNode t)
{
if (t == null)
{
return;
}
traverse(t.left);
bool inIncreasingOrder = t.val > theLast.val;
if (!inIncreasingOrder) // not in increasing order
{
if (v1 == null)
{
v1 = t; // first violation node, with bigger value
}
v2 = theLast; // second violation node, with smaller value
}
theLast = t;
traverse(t.right);
}
