Recover BST

问题描述

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

解决思路

递归思路。

中序遍历的过程中，第一个违反顺序的节点一定是错误节点的其中一个；第二个违反顺序的节点的下一个节点是另外一个错误节点。

程序

public class RecoverBST {
	private TreeNode pre = null;
	private TreeNode n1 = null;
	private TreeNode n2 = null;

	public void recoverBST(TreeNode root) {
		if (root == null) {
			return;
		}
		inorderTraversal(root);
		if (n1 != null && n2 != null) {
			int tmp = n1.val;
			n1.val = n2.val;
			n2.val = tmp;
		}
	}

	private void inorderTraversal(TreeNode root) {
		if (root==null) {
			return;
		}
		inorderTraversal(root.left);
		if (pre != null && pre.val > root.val) {
			n2 = root; // second error node‘s next
			if (n1 == null) {
				n1 = pre; // first error node
			}
		}
		pre = root;
		inorderTraversal(root.right);
	}
}