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?

中文：二叉树的前序遍历(根-左-右)。能用非递归实现吗？

递归：

public class BinaryTreePreorderTraversal {    public List<Integer> preorderTraversal(TreeNode root) {    List<Integer> list = new ArrayList<Integer>();        if(root == null)        return list;        list.add(root.val);        list.addAll(preorderTraversal(root.left));        list.addAll(preorderTraversal(root.right));        return list;    }    // Definition for binary tree     public class TreeNode {         int val;         TreeNode left;         TreeNode right;         TreeNode(int x) { val = x; }     }}

非递归：先把右节点的值压入栈中，再压入左的；弹出左的，弹出右的……。

    public List<Integer> preorderTraversal(TreeNode root){    List<Integer> list = new ArrayList<Integer>();    if(root == null)    return list;    Stack<TreeNode> stack = new Stack<TreeNode>();    stack.push(root);    while(!stack.isEmpty()){    TreeNode node = stack.pop();    list.add(node.val);      if(node.right != null)    stack.push(node.right);    if(node.left != null)    stack.push(node.left);    }    return list;    }