[leetcode] Recover Binary Search Tree

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?

思路：

一个BST树，如果中序遍历输出，那么一定是从小到大输出的。于是就可以中序遍历二叉树，如果第一次发现一个节点的值比之前访问的节点值小，那么之前的那个节点就是被交换的第一个节点。接着向后访问，如果是第二次找到某个节点的值比之前访问的节点值小，那么这个节点就是被交换的第二个节点。这是因为第一次找到的是交换顺序中比较大的，第二次是比较小的。但是有一个问题，如果两个节点相邻就会出问题，只能找到一个。仔细分析，如果交换的是两个相邻的节点，第一个节点是访问之前的节点，第二个自然是当前访问的节点。所以，不管怎样，只要满足一个节点的值比之前访问的节点值小，被交换的第二个节点一定是当前访问的节点。

题解：

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

View Code

转载于:https://www.cnblogs.com/jiasaidongqi/p/4277067.html

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