Recover Binary Search Tree

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.

 

Analyse: Traverse the binary tree inorderly, then find the two wrong numbers and swap them.

1. Recursion

    Runtime: 56ms.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode* root) {
        vector<TreeNode* > store; //store the nodes visited by indorder traversal
        if(!root || (!root->left && !root->right)) return;
        inorder(root, store);
        
        int left = 0, right = store.size() - 1;
        for(int i = 0; i < store.size(); i++){
            if(store[i + 1]->val < store[i]->val){
                left = i;
                break;
            }
        }
        for(int j = store.size() - 1; j >= 0; j--){
            if(store[j]->val < store[j - 1]->val) {
                right = j;
                break;
            }
        }
        swap(store[left]->val, store[right]->val);
        return ;
    }
    void inorder(TreeNode* root, vector<TreeNode* > &store){
        if(!root) return;
        
        inorder(root->left, store);
        store.push_back(root);
        inorder(root->right, store);
    }
};

 

2. Iteration

Runtime: 56ms.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode* root) {
        vector<TreeNode* > store; //store the nodes visited by indorder traversal
        if(!root || (!root->left && !root->right)) return;
        TreeNode* current = root;
        stack<TreeNode* > stk;
        
        while(!stk.empty() || current){
            if(current){
                stk.push(current);
                current = current->left;
            }
            else{
                current = stk.top();
                stk.pop();
                store.push_back(current);
                current = current->right;
            }
        }
        
        int left = 0, right = store.size() - 1;
        for(int i = 0; i < store.size(); i++){
            if(store[i + 1]->val < store[i]->val){
                left = i;
                break;
            }
        }
        for(int j = store.size() - 1; j >= 0; j--){
            if(store[j]->val < store[j - 1]->val) {
                right = j;
                break;
            }
        }
        swap(store[left]->val, store[right]->val);
        return ;
    }
};