题目

类型：Tree

难度：中等

解题思路

代码

class Solution {
    public void recoverTree(TreeNode root) {
        TreeNode x = null, y = null, pred = null, predecessor = null;
        while (root != null) {
            if (root.left != null) {
                // predecessor 节点就是当前 root 节点向左走一步，然后一直向右走至无法走为止
                predecessor = root.left;
                while (predecessor.right != null && predecessor.right != root) {
                    predecessor = predecessor.right;
                }
                // 让 predecessor 的右指针指向 root，继续遍历左子树
                if (predecessor.right == null) {
                    predecessor.right = root;
                    root = root.left;
                }
                // 说明左子树已经访问完了，我们需要断开链接
                else {
                    if (pred != null && root.val < pred.val) {
                        y = root;
                        if (x == null) {
                            x = pred;
                        }
                    }
                    pred = root;
                    predecessor.right = null;
                    root = root.right;
                }
            }
            // 如果没有左孩子，则直接访问右孩子
            else {
                if (pred != null && root.val < pred.val) {
                    y = root;
                    if (x == null) {
                        x = pred;
                    }
                }
                pred = root;
                root = root.right;
            }
        }
        swap(x, y);
    }
    public void swap(TreeNode x, TreeNode y) {
        int tmp = x.val;
        x.val = y.val;
        y.val = tmp;
    }
}