[LeetCode] Recover Binary Search Tree

来源：互联网发布：怎么查淘宝消费总额编辑：程序博客网时间：2024/05/08 08:47

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?

主要思路：

使用树的递归中序遍历，当出现当前值比前一个小的时候，就存在不合法的节点。用pre存中序遍历时当前节点的前一个节点，方便值的大小对比。用p,q记录两个不合法序列的位置，p指向较小的值，q指向较大的值。

/** * 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 *p, *q, *pre;    void recoverTree(TreeNode *root) {        p = q = pre = nullptr;        inorder(root);        swap(p->val, q->val);    }    void inorder(TreeNode *root)    {        if(root == nullptr)  return;        if(root->left)  inorder(root->left);        if(pre && pre->val > root->val)        {            if(p == nullptr)                p = pre;            q = root;        }        pre = root;        if(root->right) inorder(root->right);    }};

参考：http://www.cnblogs.com/tgkx1054/archive/2013/05/24/3096830.html

0 0