LeetCode OJ - Binary Tree Preorder Traversal (C++) (try 3 times) (Iterative)

為了練習尋找實習時會碰到的問題，我用別人推薦的 LeetCode online judge。
希望能直接寫程式，不用事先編譯就能直接AC。
這是下面是我寫的 Binary Tree Preorder Traversal 的解法。

To practice future interview question, I use LeetCode online judge.
I am trying to type code directly on website without compile in other tool.
The following program is my Binary Tree Preorder Traversal solution.

編譯錯誤次數 Compile Error number: 0
嘗試次數 Try number: 3 (CE: 2, AC:1)
是否事先在其他工具編譯 if compile first in other tool: No

使用的程式語言 using programming language: C++
以前是否看過 seen this problem before: Yes

CE原因:
1. 宣告Stack的時候，用stack<TreeNode> nodes; 少一個分號，應該是stack<TreeNode*> nodes;
2. 其中一行忘記加分號

這個是Iterative版本! 請看這邊有遞迴版本~
使用stack簡直易如反掌啊~~~有他真好XD

點這裡看題目 Click here to see this Problem!

	/**
	* Definition for binary tree
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };
	*/
	class Solution {
	private:
	vector<int> preOrder;
	public:
	vector<int> preorderTraversal(TreeNode *root) {
	stack<TreeNode*> nodes;
	if(root == NULL)//in case the input is empty
	return preOrder;
	nodes.push(root);
	while(!nodes.empty()){
	TreeNode *tmpNode = nodes.top();
	//psuh the root's value into a vector
	preOrder.push_back(tmpNode -> val);
	nodes.pop();//pop the tmpNode from stack
	//because statk is LIFO, the right node is latter than left node
	if(tmpNode -> right != NULL) nodes.push(tmpNode -> right);
	if(tmpNode -> left != NULL) nodes.push(tmpNode -> left);
	}
	return preOrder;
	}
	};

view raw LeetCode OJ - Binary Tree Preorder Traversal (C++) (try 3 times) (Iterative).cpp hosted with ❤ by GitHub

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