Leetcode: Recover Binary Search Tree

最新推荐文章于 2019-01-13 06:58:44 发布

原创最新推荐文章于 2019-01-13 06:58:44 发布 · 2.3k 阅读

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#Leetcode #Binary Search Tree #recursive

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本文介绍了一种解决二叉搜索树中两个元素因误操作而交换位置的问题，通过递归方法找到并交换回正确的位置，不改变树的结构。详细解释了如何在O(1)空间复杂度下实现这一目标。

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.

递归，本以为用递归会超时呢，居然没有。。。

有一点需要注意，只要当前子树中的左子树中的结点和根或者右子树中的值交换，换完之后都将是左子树中的最大值。右子树同理，是右子树中的最小值。

如果交换是左子树中的两个值交换，即交换不跨越当前根节点，就交给递归吧。

/**
 * 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* findLargestInLeft(TreeNode* root)
    {
        if(root==NULL) return NULL;
    	TreeNode* left = NULL;
		TreeNode* right = NULL;
		if(root->left)
			left = findLargestInLeft(root->left);
		if(root->right)
			right = findLargestInLeft(root->right);
		if(left && right)
		{
			TreeNode* largest = root->val > left->val ? root : left;
			largest = largest->val > right->val ? largest : right;
			return largest;
		}else if(left && right==NULL)
		{
			return root->val > left->val ? root : left;
		}else if(left==NULL && right)
		{
			return root->val > right->val ? root : right;
		}else
			return root;
    }
    TreeNode* findLeastInRight(TreeNode* root)
    {
        if(root ==NULL) return NULL;
        TreeNode* left = NULL;
		TreeNode* right = NULL;
		if(root->left != NULL)
			left = findLeastInRight(root->left);
		if(root->right)
			right = findLeastInRight(root->right);
		if(left && right)
		{
			TreeNode* least = root->val < left->val ? root : left;
			least = least->val < right->val ? least : right;
			return least;
		}else if(left && right == NULL)
		{
			return root->val < left->val ? root : left;
		}else if(left == NULL && right)
		{
			return root->val < right->val ? root : right;
		}else
			return root;
    }
    void recoverTree(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root == NULL)return;
        TreeNode* left = findLargestInLeft(root->left);
        TreeNode* right = findLeastInRight(root->right);
        if(left && right){
            if(left->val > root->val && root->val > right->val)
				swap(left->val, right->val);
			else if(root->val < left->val && left->val < right->val)
				swap(left->val, root->val);
			else if(left->val < right->val && right->val < root->val)
				swap(root->val, right->val);
			else{
				recoverTree(root->left);
				recoverTree(root->right);
			}
        }else if(left != NULL && right==NULL){
			if(root->val < left->val)
				swap(root->val, left->val);
			else
				recoverTree(root->left);
        }else if(left==NULL && right!=NULL){
			if(root->val > right->val)
				swap(root->val, right->val);
			else
				recoverTree(root->right);
        }
    }
};