标签：
Given a binary tree, return the preorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3},
1
2
/
3
return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?
Analyse: root, root->left, root->right
1. Recursion
Runtime: 0ms
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> preorderTraversal(TreeNode *root) { 13 vector<int> result; 14 preorder(root, result); 15 return result; 16 } 17 void preorder(TreeNode* root, vector<int>& result){ 18 if(root){ 19 result.push_back(root->val); 20 preorder(root->left, result); 21 preorder(root->right, result); 22 } 23 return ; 24 } 25 };
2. Iteration
Runtime: 4ms
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> preorderTraversal(TreeNode *root) { 13 vector<int> result; 14 stack<TreeNode*> stk; 15 if(root) stk.push(root); 16 else return result; 17 18 while(!stk.empty()){ 19 TreeNode* temp = stk.top(); 20 result.push_back(temp->val); 21 stk.pop(); 22 23 if(temp->right) stk.push(temp->right); 24 if(temp->left) stk.push(temp->left); 25 } 26 return result; 27 } 28 };
Binary Tree Preorder Traversal
标签：
原文地址：http://www.cnblogs.com/amazingzoe/p/4679784.html
