[LeetCode] Binary Tree Preorder Traversal, Solution-优快云博客

本文介绍了一种使用栈实现二叉树前序遍历的迭代算法，并提供了两种不同的实现方式，帮助读者理解如何避免使用递归进行树的遍历。

Given a binary tree, return the preorder traversal of its nodes’ values.
For example:
Given binary tree {1,#,2,3},

return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?
[Thoughts]
PreOrder:
1. visit node
2. visit node->left
3. visit node->right
对于递归的解法很清晰，
function preorder(root)

if root == NULL return;
print root;
preoder(node->left);
preOrder(node->right);

将该递归函数转换成非递归的话，一般都要借助于栈。

[Code]

1:  vector<int> preorderTraversal(TreeNode *root) {  
2:       stack<TreeNode*> tStack;  
3:       vector<int> result;  
4:       while(tStack.size()>0 || root != NULL)  
5:       {  
6:            if(root != NULL)  
7:            {  
8:                 result.push_back(root->val);  
9:                 if(root->right !=NULL)  
10:                 tStack.push(root->right);  
11:                 root = root->left;  
12:            }  
13:            else  
14:            {  
15:                 root = tStack.top();  
16:                 tStack.pop();  
17:            }  
18:       }  
19:       return result;  
20:  }

Update 08/20/2014. Another way to use stack

1:       vector<int> preorderTraversal(TreeNode *root) {  
2:            stack<TreeNode*> stack;  
3:            vector<int> result;  
4:            TreeNode* cur = root;  
5:            while(cur != NULL || stack.size() != 0)  
6:            {  
7:                 while(cur != NULL)  
8:                 {  
9:                      result.push_back(cur->val);  
10:                      stack.push(cur);  
11:                      cur = cur->left;  
12:                 }  
13:                 cur = stack.top();  
14:                 stack.pop();  
15:                 cur = cur->right;  
16:            }  
17:            return result;  
18:       }