树二叉查找树 99. Recover Binary Search Tree (Hard)-优快云博客

本文链接：https://blog.youkuaiyun.com/HuiFeiDeTuoNiaoGZ/article/details/120029426

答案方法一有误，暂未知

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void inorder(TreeNode* root, vector<int> &nums){
        if(root == nullptr)
            return;
        inorder(root->left, nums);
        nums.push_back(root->val);
        inorder(root->right, nums);
    }
    pair<int, int> findTwoSwapped(vector<int> &nums){
        int n = nums.size();
        int x = -1, y = -1;
        for(int i = 0; i < n-1; i++){
            if(nums[i+1] < nums[i]){
                y = nums[i+1];
                if(x == -1){
                    x = nums[i];
                }
                else break;
            }
        }
        return {x, y};
    }
    void recover(TreeNode* r, int count, int x, int y){
        if(r != nullptr){
            if(r->val == x || r->val == y){
                r->val = r->val == x ? y : x;
                if(--count == 0)
                    return;
            }
            recover(r->left, count, x, y);
            recover(r->right, count, x, y);
        }
    }
    void recoverTree(TreeNode* root) {
        vector<int> nums;
        inorder(root, nums);
        pair<int, int> swapped = findTwoSwapped(nums);
        recover(root, 2, swapped.first, swapped.second);
    }
};

隐式中序遍历

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode* root) {
        stack<TreeNode*> stk;
        TreeNode* x = nullptr;
        TreeNode* y = nullptr;
        TreeNode* pred = nullptr;

        while(!stk.empty() || root != nullptr){
            while(root != nullptr){
                stk.push(root);
                root = root->left;
            }
            root = stk.top();
            stk.pop();
            if(pred != nullptr && root->val < pred->val) {
                y = root;
                if(x == nullptr){
                    x = pred;
                }
                else break;
            }
            pred = root;
            root = root->right;
        }
        swap(x->val, y->val);
    }
};

mirrors算法

class Solution {
public:
    void recoverTree(TreeNode* root) {
        TreeNode *x = nullptr, *y = nullptr, *pred = nullptr, *predecessor = nullptr;

        while (root != nullptr) {
            if (root->left != nullptr) {
                // 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 != nullptr && 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 == nullptr) {
                    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 != nullptr && root->val < pred->val) {
                        y = root;
                        if (x == nullptr) {
                            x = pred;
                        }
                    }
                    pred = root;

                    predecessor->right = nullptr;
                    root = root->right;
                }
            }
            // 如果没有左孩子，则直接访问右孩子
            else {
                if (pred != nullptr && root->val < pred->val) {
                    y = root;
                    if (x == nullptr) {
                        x = pred;
                    }
                }
                pred = root;
                root = root->right;
            }
        }
        swap(x->val, y->val);
    }
};