【Leetcode】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?

二叉查找树中有两个节点被调换了，要求在不改变树的结构的情况下恢复原状。

思路：找到第一个比它的直接后继结点中的值更大的结点A，然后向后找到【第一个】比A中的值大的结点B，然后将A的值和B的直接前驱中的值交换，返回。

1、空间复杂度O(n)的做法

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:vector<TreeNode*> com;void Traverse(TreeNode *root){if (!root) return;Traverse(root->left);com.push_back(root);Traverse(root->right);}    void recoverTree(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() functioncom.clear();    Traverse(root);TreeNode *itema=NULL,*itemb=NULL;for (size_t i=0; i<com.size(); ++i){if (!itema && i+1 < com.size() && com[i]->val > com[i+1]->val){itema = com[i];while (++i < com.size()){if (com[i] > itema->val)break;}if (i == com.size())itemb = com.back();elseitemb = com[i-1];int temp = itema->val;itema->val = itemb->val;itemb->val = temp;                                break;}}    }};

2 、空间复杂度O(1)的做法

/** * 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 *prev;TreeNode *itema, *itemb;void traverse(TreeNode *root){if(!root) return;traverse(root->left);if(prev){if(itema && root->val > itema->val)itemb = prev;if(!itema && prev->val > root->val)itema = prev;// pay attention!if (itema && itemb)return;}prev = root;traverse(root->right);}       void recoverTree(TreeNode *root) {prev = itema = itemb = NULL;traverse(root);if (itema && !itemb)itemb = prev;if (itema && itemb){int temp = itema->val;itema->val = itemb->val;itemb->val = temp;}    }};