题目

方法

利用BST中序遍历是一个有序的序列。

中序遍历BST，遇到顺序错误的则标记，找到两个错误，交换其值即可。

若只找到一个，则说明第一个标记的后面的结点，是错误的。

    public void recoverTree(TreeNode root) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode node = root;
        TreeNode first = null;
        TreeNode second = null;
        TreeNode last = null;
        while (node != null || !stack.isEmpty()) {
            while(node != null) {
                stack.push(node);
                node = node.left;
            }
            TreeNode curNode = stack.pop();
            if (last == null) {
                last = curNode;
            } else {
                if (last.val > curNode.val) {
                    if (first == null) {
                        first = last;
                        second = curNode;
                    } else {
                        second = curNode;
                        break;
                    }
                }
                last = curNode;
            }
            node = curNode.right;
        }
        
        int temp = first.val;
        first.val = second.val;
        second.val = temp;
    }

Recover Binary Search Tree,布布扣,bubuko.com

Recover Binary Search Tree

原文：http://blog.csdn.net/u010378705/article/details/28672765