[Leetcode][第99题][JAVA][恢复二叉搜索树][中序遍历]

【问题描述】[困难]

【解答思路】

1. 显示中序遍历

时间复杂度：O(N) 空间复杂度：O(N)

class Solution {
    public void recoverTree(TreeNode root) {
        List<Integer> nums = new ArrayList<Integer>();
        inorder(root, nums);
        int[] swapped = findTwoSwapped(nums);
        recover(root, 2, swapped[0], swapped[1]);
    }

    public void inorder(TreeNode root, List<Integer> nums) {
        if (root == null) {
            return;
        }
        inorder(root.left, nums);
        nums.add(root.val);
        inorder(root.right, nums);
    }

    public int[] findTwoSwapped(List<Integer> nums) {
        int n = nums.size();
        //第一次找到取前面 第二次找到取后面
        int x = -1, y = -1;
        for (int i = 0; i < n - 1; ++i) {
            if (nums.get(i + 1) < nums.get(i)) {
                y = nums.get(i + 1);
                if (x == -1) {
                    x = nums.get(i);
                } else {
                    break;
                }
            }
        }
        return new int[]{x, y};
    }

    public void recover(TreeNode root, int count, int x, int y) {
        if (root != null) {
            if (root.val == x || root.val == y) {
                root.val = root.val == x ? y : x;
                if (--count == 0) {
                    return;
                }
            }
            recover(root.right, count, x, y);
            recover(root.left, count, x, y);
        }
    }
}

2. 隐式中序遍历 栈

时间复杂度：O(N) 空间复杂度：O(H)

class Solution {
    public void recoverTree(TreeNode root) {
        Deque<TreeNode> stack = new ArrayDeque<TreeNode>();
        TreeNode x = null, y = null, pred = null;

        while (!stack.isEmpty() || root != null) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            if (pred != null && root.val < pred.val) {
                y = root;
                if (x == null) {
                    x = pred;
                } else {
                    break;
                }
            }
            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;
    }
}

3. Morris 中序遍历

时间复杂度：O(N) 空间复杂度：O(1)

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;
    }
}

【总结】

1. 交换的思想

两个指针或两个标志找逆序

2. 二叉搜索树 中序遍历 单调递增

3.空间复杂度优化 Morris 中序遍历优化 空间复杂度：O(1)

转载链接：https://leetcode-cn.com/problems/recover-binary-search-tree/solution/hui-fu-er-cha-sou-suo-shu-by-leetcode-solution/