Recover Binary Search Tree

最新推荐文章于 2022-03-13 20:36:15 发布

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最新推荐文章于 2022-03-13 20:36:15 发布

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分类专栏： C++算法文章标签：二叉树

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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.

为了在空间O(1)内完成，我们设置了3个指针。pre为当前节点的前序节点，n1和n2为待交换的节点。我们在递归中用中序遍历的方式，因为二叉搜索树按中序遍历是一个有序递增序列。找到当前节点大于右子树根节点的错误情况，并将值付给n1和n2，找到错误的节点只需要最后将其交换即可。

/**
 * 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) {
        TreeNode *n1=NULL, *n2=NULL, *pre=NULL;
        find(root,pre,n1,n2);
        int tmp=n1->val;
        n1->val=n2->val;
        n2->val=tmp;
    }

    void find(TreeNode *root, TreeNode *&pre, TreeNode *&n1, TreeNode *&n2){
        if(root==NULL)
            return;
        find(root->left,pre,n1,n2);
        if(pre&&pre->val>root->val){  //前序节点大于右子树根节点
            if(n1==NULL)
                n1=pre;
            n2=root;
        }
        pre=root;
        find(root->right,pre,n1,n2);
    }
};