Flatten Binary Tree to Linked List @leetcode
=看完别人的代码，就无限接近于认为自己是沙茶。大家的思路怎么这么简洁，以至于我能看懂，我估计我的代码别人都看不懂。
1 void flatten(TreeNode *root) { 2 // IMPORTANT: Please reset any member data you declared, as 3 // the same Solution instance will be reused for each test case. 4 TreeNode*tmp; 5 flattenpiece(root,tmp); 6 } 7 void flattenpiece(TreeNode *root,TreeNode *&tail){ 8 if(root == NULL) 9 return; 10 TreeNode*leftTreeRoot = root->left; 11 TreeNode*rightTreeRoot = root->right; 12 if(leftTreeRoot == NULL) 13 { 14 TreeNode*tmp; 15 flattenpiece(rightTreeRoot,tmp); 16 if(rightTreeRoot == NULL) 17 tail = root; 18 else 19 tail =tmp; 20 return; 21 } 22 //TreeNode*leftTreeTail; 23 flattenpiece(leftTreeRoot,tail); 24 root->right = leftTreeRoot; 25 root->left = NULL; 26 if(tail) 27 { 28 if(rightTreeRoot) 29 { 30 tail->right = rightTreeRoot; 31 tail->left = NULL; 32 tail = rightTreeRoot; 33 } 34 } 35 flattenpiece(rightTreeRoot,tail); 36 } 37 }
=思路嗷嗷清晰的人的代码：
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 // IMPORTANT: Please reset any member data you declared, as 14 // the same Solution instance will be reused for each test case. 15 if (root == NULL) return; 16 if (root->left == NULL && root->right == NULL) { 17 return; 18 } else if (root->left == NULL) { 19 flatten(root->right); 20 } else if (root->right == NULL) { 21 root->right = root->left; 22 root->left = NULL; 23 flatten(root->right); 24 } else { 25 flatten(root->left); 26 flatten(root->right); 27 TreeNode* p = root->left; 28 while (p->right) { 29 p = p->right; 30 } 31 p->right = root->right; 32 root->right = root->left; 33 root->left = NULL; 34 } 35 } 36 };
=我是弧度
