99. Recover Binary Search Tree (Tree; DFS)

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?

法I：BST的中序遍历结果是递增序列。把这个递增序列存储到数组中，再遍历一遍数组，便可知道swap的两个元素

class Solution {
public:void recoverTree(TreeNode *root) {vals.clear();treeNodes.clear();inorderTraverse(root);sort(vals.begin(), vals.end());for (int i = 0; i < treeNodes.size(); ++i){treeNodes[i]->val = vals[i]; //只改变值，不改结构
        }}void inorderTraverse(TreeNode* root){if (!root)return;inorderTraverse(root->left);vals.push_back(root->val);treeNodes.push_back(root);inorderTraverse(root->right);        }
private:vector<int> vals;  vector<TreeNode*> treeNodes;
};

法II：空间复杂度O(1)的解决方案。在中序遍历的时候，当第一次扫描到前一个节点>后一个节点，那么必定是前一个节点错了；当第二次扫描到前一个节点>后一个节点，那么必定是后一个节点错了。有一种特殊情况是：第一次扫描到错误的时候，两个节点都是在错误的位置，所以在第一次扫描到错误时，也需要把后一个节点记录下来。

/*** 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* pre = NULL;TreeNode* swap1 = NULL;TreeNode* swap2 = NULL;inOrderTraverse(root,pre,swap1,swap2);int tmp = swap1->val;swap1->val = swap2->val;swap2->val = tmp;}void inOrderTraverse(TreeNode* root, TreeNode* &pre,TreeNode* &swap1,TreeNode* &swap2){ //important to use &, otherwise new object will use a new address and the result won't bring back to caller//visit left childif(root->left) inOrderTraverse(root->left,pre,swap1,swap2);//visit rootif(pre && root->val < pre->val){swap2 = root;if(swap1==NULL) {swap1 = pre;}}pre = root;//visit right childif(root->right) inOrderTraverse(root->right,pre,swap1,swap2);}
};

转载于:https://www.cnblogs.com/qionglouyuyu/p/4854218.html

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