Leetcode: 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, could you do it iteratively?

Analysis: 第一反应肯定是recursion(递归), 非常直接，但是题目要求不能用递归。如果要使用迭代的方法来写preorder traversal，最大的问题是要如何确定遍历节点的顺序，因为树的pre-order traversal其实很类似图的DFS，DFS可以用Stack来写，所以这里写pre-order traversal也可以用stack来实现迭代的写法。

Iterative: 其实就是用一个栈来模拟递归的过程。所以算法时间复杂度也是O(n)，空间复杂度是栈的大小O(logn)。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        TreeNode p = root;
        Stack<TreeNode> stack = new Stack<>();
        while (p!=null || !stack.isEmpty()) {
            if (p != null) {
                stack.push(p);
                res.add(p.val);
                p = p.left;
            }
            else {
                TreeNode node = stack.pop();
                p = node.right;
            }
        }
        return res;
    }
}

 

recursion:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        preorder(root, result);
        return result;
    }
    
    public void preorder(TreeNode root, ArrayList<Integer> result) {
        if (root == null) return;
        result.add(root.val);
        preorder(root.left, result);
        preorder(root.right, result);
    }
}