Leetcode recover Binary Search Tree

class Solution {private:    vector<TreeNode*> nodes;public:    void recoverTree(TreeNode *root) {        nodes.clear();        dfs(root);        // 1 5 3 4 2 6 7        // 1 3 2 4        // 3 2        int len = nodes.size();        if (len < 1) return;                int pre = nodes[0]->val;        int cur = 0;        int first = -1;        int second= -1;        for (int i=1; i<len; i++) {            cur = nodes[i]->val;            if (cur < pre) {                if (first < 0) {                    first = i-1;                } else {                    second= i;                }            }            pre = cur;        }        if (second < 0) second = first + 1;        int t = nodes[first]->val;        nodes[first]->val = nodes[second]->val;        nodes[second]->val= t;    }        void dfs(TreeNode* root) {        if (root == NULL) return;        dfs(root->left);        nodes.push_back(root);        dfs(root->right);    }};

First let's get a "A solution using O (N) Space is pretty straight forward "method. If the recursive call stack does not involve any extra space, you can simply traverse the stack in sequence and modify it. What if so?

For a pseudo-constant space:

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {private:    vector<TreeNode*> nodes;    TreeNode* pre;    TreeNode* first;    TreeNode* second;public:    void recoverTree(TreeNode *root) {        nodes.clear();        pre = first = second = NULL;        dfs(root);        // case a. 1 5 3 4 2 6 7        // case b. 1 3 2 4        // case c. 3 2        // case d. 3        // case e. NULL        if (second == NULL) return; // case (d,e)        int t = first->val;        first->val = second->val;        second->val= t;    }        void dfs(TreeNode* root) {        if (root == NULL) return;        dfs(root->left);        visit(root);        dfs(root->right);    }        void visit(TreeNode* node) {        if (pre == NULL) {            pre = node;            return;        }        if (node->val < pre->val) {            if (first == NULL) {                first = pre;                second= node;   // assume swap with node next to pre(case b,c)            } else {                second= node;   // fix above assumption(case a)            }        }        pre = node;    }};

 

Leetcode recover Binary Search Tree