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There are three methods to do it:
1. recursive(use memory stack): (Time O(n), Space O(logn)
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 getTree(TreeNode *root, vector<int> &result) { 13 if (!root) return; 14 getTree(root->left, result); 15 result.push_back(root->val); 16 getTree(root->right, result); 17 } 18 vector<int> inorderTraversal(TreeNode *root) { 19 vector<int> result; 20 getTree(root, result); 21 return result; 22 } 23 };
2. directly use stack: Same with previous
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 vector<int> inorderTraversal(TreeNode *root) { 13 vector<int> result; 14 if (!root) return result; 15 stack<TreeNode *> s; 16 while (root || !s.empty()) { 17 if (root) { 18 s.push(root); 19 root = root->left; 20 } else { 21 result.push_back(s.top()->val); 22 root = s.top()->right; 23 s.pop(); 24 } 25 } 26 return result; 27 } 28 };
3. Two pointers (Time O(n), Space O(1)):
This method is trying to temporarily change the tree structure as a so called "linkedlist".
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 vector<int> inorderTraversal(TreeNode *root) { 13 vector<int> result; 14 if (!root) return result; 15 TreeNode *prev = NULL; 16 while (root) { 17 if (!root->left) { 18 result.push_back(root->val); 19 root = root->right; 20 } else { 21 prev = root->left; 22 while(prev->right && prev->right != root) prev = prev->right; 23 if (!prev->right) { 24 prev->right = root; 25 root = root->left; 26 } else { 27 prev->right = NULL; 28 result.push_back(root->val); 29 root = root->right; 30 } 31 } 32 } 33 return result; 34 } 35 };
LeetCode – Refresh – Binary Tree Inorder Traversal
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原文地址:http://www.cnblogs.com/shuashuashua/p/4346169.html