144. Binary Tree Preorder Traversal

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

For example:
Given binary tree {1,#,2,3},

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2
3
4
5
1
 \
  2
 /
3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

解法1：Iterative version 用stack

递归的就不写了，实在太简单。
用stack的方法的关键点在于，每一条路劲在探寻最左元素的时候，要把所有的parent node都push到stack中。
如果当前的路径到头了之后，从stack里pop出上一个parent，然后移步到右边。
注意终止条件是需要node！= null同时stack不为空。
C++

1

Java

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/**
 * 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<Integer>();
        if (root == null) {
            return res;
        }
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode current = root;
        while (current != null || !stack.isEmpty()) {
            // push all left nodes into stack
            while (current != null) {
                res.add(current.val);
                stack.push(current);
                current = current.left;
            }
            if (!stack.isEmpty()) {
                TreeNode temp = stack.pop();
                current = temp.right;   // move to the right cell
            }
        }
        return res;
    }
}