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[LintCode/LeetCode] Binary Tree Preorder Traversal

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摘要：當(dāng)你被的時候，人家問，一定要說，當(dāng)然可以啦所以，不用，怎么辦幸好，這是一道的題目，只要逐層遍歷，就可以了。所以，試一試用堆棧的做法吧記得的特點是后入先出哦

Problem

Given a binary tree, return the preorder traversal of its nodes" values.

Example

Given:

    1
   / 
  2   3
 / 
4   5

return [1,2,4,5,3].

Challenge

Can you do it without recursion?

Note

當(dāng)你被challenge的時候，人家問，Can you do it without recursion? 一定要說，當(dāng)然可以啦！所以，不用recursion，怎么辦？幸好，這是一道preorder的題目，只要逐層遍歷，就可以了。所以，試一試用堆棧的做法吧！記得Stack的特點是后入先出哦！

Solution

Recursion

public class Solution {
    public ArrayList preorderTraversal(TreeNode root) {
        ArrayList res = new ArrayList();
        if (root == null) return res;
        helper(res, root);
        return res;
    }
    public void helper(ArrayList res, TreeNode root) {
        res.add(root.val);
        if (root.left != null) helper(res, root.left);
        if (root.right != null) helper(res, root.right);
    }
}

Stack

public class Solution {
    public ArrayList preorderTraversal(TreeNode root) {
        ArrayList res = new ArrayList();
        Stack stack = new Stack();
        if (root == null) return res;
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            res.add(cur.val);
            if (cur.right != null) stack.push(cur.right);
            if (cur.left != null) stack.push(cur.left);
        }
        return res;
    }
}

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