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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?
思路：中序递归的思路。用pre记录前一个节点，找出违反二叉搜索树规则的节点。中序遍历序列中，第一次违反二叉搜索树规则的节点的前一个节点是要修改的节点。第二次违反二叉搜索树规则的节点本身是要修改的节点
时间复杂度O(n),空间复杂度O(1)
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 TreeNode *pre = NULL; 13 TreeNode *mistake_node1 = NULL; 14 TreeNode *mistake_node2 = NULL; 15 16 void InorderTraversal(TreeNode *root) { 17 if (root != NULL) { 18 InorderTraversal(root->left); 19 20 if (pre != NULL && root->val < pre->val) { 21 if (!mistake_node1) { 22 mistake_node1 = pre; 23 mistake_node2 = root; 24 } else { 25 mistake_node2 = root; 26 } 27 } 28 pre = root; 29 InorderTraversal(root->right); 30 } 31 } 32 void recoverTree(TreeNode *root) { 33 InorderTraversal(root); 34 if (mistake_node1 && mistake_node2) 35 swap(mistake_node1->val, mistake_node2->val); 36 } 37 };
[LeetCode] Recover Binary Search Tree
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原文地址：http://www.cnblogs.com/vincently/p/4240253.html
