[LeetCode] Binary Tree Preorder Traversal (non-recursive first order Traversal)

[LeetCode] Binary Tree Preorder Traversal (non-recursive first order Traversal)

 

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, cocould you do it iteratively

Solution:

The forward traversal of a binary tree. A non-recursive method is required. Let's take a look at the solution of recursive methods.

1. 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
 
   preorderTraversal(TreeNode* root) {        vector
  
    result;        preorderHelper(result, root);        return result;    }        void preorderHelper(vector
   
    & result, TreeNode* root){        if(root==NULL){            return;        }        result.push_back(root->val);        preorderHelper(result, root->left);        preorderHelper(result, root->right);    }};
   
  
 

2. Non-recursive Solution

 

We can use two data structures to store intermediate states. Use a queue to store left children and stack to store right children. Traverse all left children first.

 

/** * 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
 
   preorderTraversal(TreeNode* root) {        vector
  
    result;        queue
   
     l;        stack
    
      r;        if(root!=NULL){            l.push(root);        }        while(!l.empty()||!r.empty()){            TreeNode* node = NULL;            if(!l.empty()){                node=l.front();                l.pop();            }else{                node=r.top();                r.pop();            }            result.push_back(node->val);            if(node->left!=NULL){                l.push(node->left);            }            if(node->right!=NULL){                r.push(node->right);            }        }                return result;    }};