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?
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 if(root==NULL) 14 return v; 15 v.push_back(root->val); 16 preorderTraversal(root->left); 17 preorderTraversal(root->right); 18 return v; 19 } 20 vector<int> v; 21 };
非递归
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> v; 14 if(root==NULL) 15 return v; 16 stack<TreeNode *> tree; 17 tree.push(root); 18 while(!tree.empty()){ 19 TreeNode *p=tree.top(); 20 v.push_back(p->val); 21 tree.pop(); 22 if(p->right!=NULL) 23 tree.push(p->right); 24 if(p->left!=NULL) 25 tree.push(p->left); 26 } 27 return v; 28 } 29 };