[Leetcode] [tree] [flatten binary tree to linked list]

Link the tree with right in the order of traversal in the forward order.

I thought for a long time, but I didn't have any idea at all. I always thought that the inplace solution was very clever.

Later, I decided to use a variable to record the current access point. It was so witty ~~

 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     void flatten(TreeNode *root) {13         now = NULL;14         findRes(root);15     }16 private:17     TreeNode * now;18     void findRes(TreeNode *root) {19         if (root == NULL) {20             return;21         }22         findRes(root->right);23         findRes(root->left);24         root->left = NULL;25         root->right = now;26         now = root;27     }28 };

 

After an AC, I read the question and found that the ideas differ greatly! The version of iteration is clear. Once again, it is also an AC.

 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     void flatten(TreeNode *root) {13         if (root == NULL) {14             return;15         }16         stack<TreeNode*> s;17         s.push(root);18         while (!s.empty()) {19             TreeNode * now = s.top();20             s.pop();21             if (now->right != NULL) {22                 s.push(now->right);23             }24             if (now->left != NULL) {25                 s.push(now->left);26             }27             now->left = NULL;28             if (!s.empty()) {29                 now->right = s.top();30             }31         }32     }33 };