[leedcode 144] 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].

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    List<Integer> res;
    public List<Integer> preorderTraversal(TreeNode root) {
     /*   //解法一：借助栈数据结构
        List<Integer> res=new ArrayList<Integer>();
        if(root==null) return res;
        Stack<TreeNode> stack=new Stack<TreeNode>();
        TreeNode node=root;
        while(node!=null||!stack.isEmpty()){//条件
            while(node!=null){
                res.add(node.val);///先序
                stack.push(node);
                node=node.left;

            }
            node=stack.pop();
            //res.add(node.val);中序
            node=node.right;
        }
        return res;*/
        
       /*解法二：
       res=new ArrayList<Integer>();
        preorder(root);
        return res;
 */
 
    //解法三：此种解法注意push子节点的顺序，先右节点，再左节点
    List<Integer> res=new ArrayList<Integer>();
    Stack<TreeNode> stack=new Stack<TreeNode>();
    if(root==null) return res;
    stack.push(root);
    while(!stack.isEmpty()){
        TreeNode node=stack.pop();
        res.add(node.val);
        if(node.right!=null){
            stack.push(node.right);
        }
        if(node.left!=null){
            stack.push(node.left);
        }
        
    }
    return res;
    }
    /*解法二：
    public void preorder(TreeNode node){
        if(node==null) return;
        res.add(node.val);
        preorder(node.left);
        preorder(node.right);
    }*/
}