Binary Tree Preorder Traversal @LeetCode
PROBLEM :
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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SIMPLE c++ IMPLEMENTATION : (Recursive Solution)
--------------------------------------------------------------------------------
/**
* 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> vec ;
if(root==NULL)
return vec ;
preorder(root,vec) ;
return vec ;
}
void preorder(TreeNode* root,vector<int> &vec){
if(root==NULL)
return ;
vec.push_back(root->val) ;
preorder(root->left,vec) ;
preorder(root->right,vec) ;
}
};
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SIMPLE c++ IMPLEMENTATION : (Iterative Solution)
--------------------------------------------------------------------------------
/**
* 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> vec ;
if(root==NULL)
return vec ;
stack<TreeNode*> sta ;
sta.push(root) ;
while(!sta.empty()){
TreeNode* curr=sta.top() ;
sta.pop() ;
vec.push_back(curr->val) ;
if(curr->right)
sta.push(curr->right) ;
if(curr->left)
sta.push(curr->left) ;
}
return vec ;
}
};
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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?
--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION : (Recursive Solution)
--------------------------------------------------------------------------------
/**
* 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> vec ;
if(root==NULL)
return vec ;
preorder(root,vec) ;
return vec ;
}
void preorder(TreeNode* root,vector<int> &vec){
if(root==NULL)
return ;
vec.push_back(root->val) ;
preorder(root->left,vec) ;
preorder(root->right,vec) ;
}
};
--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION : (Iterative Solution)
--------------------------------------------------------------------------------
/**
* 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> vec ;
if(root==NULL)
return vec ;
stack<TreeNode*> sta ;
sta.push(root) ;
while(!sta.empty()){
TreeNode* curr=sta.top() ;
sta.pop() ;
vec.push_back(curr->val) ;
if(curr->right)
sta.push(curr->right) ;
if(curr->left)
sta.push(curr->left) ;
}
return vec ;
}
};
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