[LeetCode] Recover Binary Search Tree 复原二叉搜索树

 

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?

 

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

 

这道题要求我们复原一个二叉搜索树,说是其中有两个的顺序被调换了,题目要求上说O(n)的解法很直观,这种解法需要用到递归,用中序遍历树,并将所有节点存到一个一维向量中,把所有节点值存到另一个一维向量中,然后对存节点值的一维向量排序,在将排好的数组按顺序赋给节点。这种最一般的解法可针对任意个数目的节点错乱的情况,这里先贴上此种解法:

 

// O(n) space complexity
class Solution {
public:
    void recoverTree(TreeNode *root) {
        vector<TreeNode*> list;
        vector<int> vals;
        inorder(root, list, vals);
        sort(vals.begin(), vals.end());
        for (int i = 0; i < list.size(); ++i) {
            list[i]->val = vals[i];
        }
    }
    void inorder(TreeNode *root, vector<TreeNode*> &list, vector<int> &vals) {
        if (!root) return;
        inorder(root->left, list, vals);
        list.push_back(root);
        vals.push_back(root->val);
        inorder(root->right, list, vals);
    }
};

 

然后我上网搜了许多其他解法,看到另一种是用双指针来代替一维向量的,但是这种方法用到了递归,也不是O(1)空间复杂度的解法,这里需要三个指针,first,second分别表示第一个和第二个错乱位置的节点,pre指向当前节点的中序遍历的前一个节点。这里用传统的中序遍历递归来做,不过再应该输出节点值的地方,换成了判断pre和当前节点值的大小,如果pre的大,若first为空,则将first指向pre指的节点,把second指向当前节点。这样中序遍历完整个树,若first和second都存在,则交换它们的节点值即可。这个算法的空间复杂度仍为O(n),n为树的高度,代码如下:

 

// Still O(n) space complexity
class Solution {
public:
    TreeNode *pre;
    TreeNode *first;
    TreeNode *second;
    void recoverTree(TreeNode *root) {
        pre = NULL;
        first = NULL;
        second = NULL;
        inorder(root);
        if (first && second) swap(first->val, second->val);
    }
    void inorder(TreeNode *root) {
        if (!root) return;
        inorder(root->left);
        if (!pre) pre = root;
        else {
            if (pre->val > root->val) {
                if (!first) first = pre;
                second = root;
            }
            pre = root;
        }
        inorder(root->right);
    }
};

 

这道题的真正符合要求的解法应该用的Morris遍历,这是一种非递归且不使用栈,空间复杂度为O(1)的遍历方法,可参见我之前的博客Binary Tree Inorder Traversal 二叉树的中序遍历,在其基础上做些修改,加入first, second和parent指针,来比较当前节点值和中序遍历的前一节点值的大小,跟上面递归算法的思路相似,代码如下:

 

// Now O(1) space complexity
class Solution {
public:
    void recoverTree(TreeNode *root) {
        TreeNode *first = NULL, *second = NULL, *parent = NULL;
        TreeNode *cur, *pre;
        cur = root;
        while (cur) {
            if (!cur->left) {
                if (parent && parent->val > cur->val) {
                    if (!first) first = parent;
                    second = cur;
                }
                parent = cur;
                cur = cur->right;
            } else {
                pre = cur->left;
                while (pre->right && pre->right != cur) pre = pre->right;
                if (!pre->right) {
                    pre->right = cur;
                    cur = cur->left;
                } else {
                    pre->right = NULL;
                    if (parent->val > cur->val) {
                        if (!first) first = parent;
                        second = cur;
                    }
                    parent = cur;
                    cur = cur->right;
                }
            }
        }
        if (first && second) swap(first->val, second->val);
    }
};

 

LeetCode All in One 题目讲解汇总(持续更新中...)

 

你可能感兴趣的:(Binary search)