【Leetcode】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?

二叉查找樹中有兩個節點被調換了，要求在不改變樹的結構的情況下恢復原狀。

思路：找到第一個比它的直接後繼結點中的值更大的結點A，然後向後找到【第一個】比A中的值大的結點B，然後將A的值和B的直接前驅中的值交換，返回。

1、空間複雜度O(n)的做法

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
	vector<TreeNode*> com;
	
	void Traverse(TreeNode *root)
	{
		if (!root) return;
		Traverse(root->left);
		com.push_back(root);
		Traverse(root->right);
	}
    void recoverTree(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
		com.clear();    
		Traverse(root);
		TreeNode *itema=NULL,*itemb=NULL;
		for (size_t i=0; i<com.size(); ++i)
		{
			if (!itema && i+1 < com.size() && com[i]->val > com[i+1]->val)
			{
				itema = com[i];
				while (++i < com.size())
				{
					if (com[i] > itema->val)
						break;
				}
				if (i == com.size())
					itemb = com.back();
				else
					itemb = com[i-1];

				int temp = itema->val;
				itema->val = itemb->val;
				itemb->val = temp;
                                break;
			}
		}
    }
};

2 、空間複雜度O(1)的做法

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution 
{
public:
	TreeNode *prev;
	TreeNode *itema, *itemb;

	void traverse(TreeNode *root)
	{
		if(!root) return;
		traverse(root->left);
		if(prev)
		{
			if(itema && root->val > itema->val)
				itemb = prev;
			if(!itema && prev->val > root->val)
				itema = prev;
			// pay attention!
			if (itema && itemb)
				return;
		}
		prev = root;
		traverse(root->right);
	}
       void recoverTree(TreeNode *root) 
	{
		prev = itema = itemb = NULL;
		traverse(root);
		if (itema && !itemb)
			itemb = prev;
		if (itema && itemb)
		{
			int temp = itema->val;
			itema->val = itemb->val;
			itemb->val = temp;
		}
    }
};