LeetCode

题目：

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?

思路：

中序遍历

package bst;

public class RecoverBinarySearchTree {
    
    private TreeNode prev = null;
    
    private TreeNode node1 = null;
    
    private TreeNode node2 = null;
    
    public void recoverTree(TreeNode root) {
        recoverTreeInternal(root);
        swap(node1, node2);
    }
    
    private void recoverTreeInternal(TreeNode root) {
        if (root == null) return;
        recoverTreeInternal(root.left);
        if (prev != null) {
            if (prev.val >= root.val) {
                if (node1 == null) 
                    node1 = prev;
                node2 = root;
            }
        }
        
        prev = root;
        recoverTreeInternal(root.right);        
    }
    
    private void swap(TreeNode node1, TreeNode node2) {
        if (node1 != null && node2 != null) {
            int tmp = node1.val;
            node1.val = node2.val;
            node2.val = tmp;
        }
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        RecoverBinarySearchTree r = new RecoverBinarySearchTree();
        TreeNode root = new TreeNode(0);
        TreeNode right = new TreeNode(1);
        root.left = right;
        r.recoverTree(root);
    }

}