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Given a binary tree, return the inorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3},
1
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
Analyse: root->left, root, root->right.
1. Recursion
Runtime: 0ms.
1 /** 2 * Definition for a binary tree node. 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 16 inorder(root, result); 17 return result; 18 } 19 void inorder(TreeNode* root, vector<int>& result){ 20 if(root->left) inorder(root->left, result); 21 result.push_back(root->val); 22 if(root->right) inorder(root->right, result); 23 } 24 };
2. Iteration
Runtime: 0ms.
1 /** 2 * Definition for a binary tree node. 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* > stk; 16 17 while(root || !stk.empty()){ 18 if(root){ 19 stk.push(root); 20 root = root->left; 21 } 22 else{ 23 root = stk.top(); 24 result.push_back(root->val); 25 stk.pop(); 26 root = root->right; 27 } 28 } 29 return result; 30 } 31 };
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原文地址：http://www.cnblogs.com/amazingzoe/p/4679898.html
