144. Binary Tree Preorder Traversal

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?

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/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
vector<int> v1;
if(root==NULL) return v1;
pretracel(root,v1);
return v1;
}
void pretracel(TreeNode* root, vector<int>&v1)
{
if(root==NULL)
return ;
v1.push_back(root->val);
if(root->left) pretracel(root->left,v1);
if(root->right) pretracel(root->right,v1);
return ;
}
};

非递归写法：

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
vector<int> v1;
pretracel(root,v1);
return v1;
}
void pretracel(TreeNode* root, vector<int>&v1)
{
stack<TreeNode*> t1;
while(!t1.empty()||root)
{
while(root)
{
t1.push(root);
v1.push_back(root->val);
root=root->left;
}
if(!t1.empty())
{
TreeNode* temp=t1.top();
t1.pop();
root=temp->right;
}
}
return ;
}
};