Binary Tree Preorder Traversal 二叉树的先序@LeetCode

用递归和迭代求Pre-Order的经典题，什么时候要把In-Order, Post-Order, Level-Order 的遍历都写一遍，因为太经典了，最好能直接背下来。

package Level3;import java.util.ArrayList;import java.util.Stack;/** *  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    /   3return [1,2,3].Note: Recursive solution is trivial, could you do it iteratively? */import Utility.TreeNode;public class S128 {public static void main(String[] args) {TreeNode root = new TreeNode(1);TreeNode n1 = new TreeNode(2);root.right = n1;System.out.println(preorderTraversal(root));System.out.println(iter(root));}public static ArrayList<Integer> preorderTraversal(TreeNode root) {ArrayList<Integer> ret = new ArrayList<Integer>();rec(root, ret);        return ret;    }public static void rec(TreeNode root, ArrayList<Integer> ret){if(root == null){return;}ret.add(root.val);rec(root.left, ret);rec(root.right, ret);}// 迭代法public static ArrayList<Integer> iter(TreeNode root) {ArrayList<Integer> ret = new ArrayList<Integer>();Stack<TreeNode> stack = new Stack<TreeNode>();if(root == null){return ret;}stack.push(root);TreeNode cur = null;while(!stack.isEmpty()){cur = stack.peek();// 保存栈顶元素ret.add(cur.val);stack.pop();// 要先放右元素再放左元素，这样才能在pop时先pop左元素再pop右元素if(cur.right != null){stack.push(cur.right);}if(cur.left != null){stack.push(cur.left);}}return ret;}}

/** * 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> ret = new ArrayList<Integer>();        if(root == null){            return ret;        }        Stack<TreeNode> stack = new Stack<TreeNode>();        stack.push(root);        while(!stack.isEmpty()){            TreeNode cur = stack.pop();            ret.add(cur.val);            if(cur.right != null)            stack.push(cur.right);            if(cur.left != null)            stack.push(cur.left);        }        return ret;    }}