标签：int node binary ret state one ini span and
Problem Statement
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?
solution one:
traverse:
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 void preorderTraverse(TreeNode *node, vector<int>& preorder){ 13 if(node == NULL){ 14 return; 15 } 16 17 preorder.push_back(node->val); 18 preorderTraverse(node->left, preorder); 19 preorderTraverse(node->right, preorder); 20 return; 21 } 22 23 vector<int> preorderTraversal(TreeNode* root) { 24 vector<int> preorder; 25 preorderTraverse(root, preorder); 26 return preorder; 27 } 28 };
Divide and conquer:
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 // this version is divide && conquer 13 vector<int> preorderTraversal(TreeNode* root) { 14 vector<int> preorder; 15 if(root == NULL){ 16 return preorder; 17 } 18 // divide 19 vector<int> left = preorderTraversal(root->left); 20 vector<int> right = preorderTraversal(root->right); 21 // conquer 22 preorder.push_back(root->val); 23 preorder.insert(preorder.end(), left.begin(), left.end()); 24 preorder.insert(preorder.end(), right.begin(), right.end()); 25 return preorder; 26 } 27 };
144. Binary Tree Preorder Traversal
标签：int node binary ret state one ini span and
原文地址：http://www.cnblogs.com/wdw828/p/6388579.html
