leetcode 144. Binary Tree Preorder Traversal

原文链接http://www.myexception.cn/program/1958647.html

题目

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

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

return [1,2,3].

解1（递归）

//前序遍历     递归        public static void preorderTraversal(TreeNode root){            if(root==null) return ;            System.out.print(root.val+"  ");            inorderTraversal(root.left);                    inorderTraversal(root.right);                                   }

解2（非递归方法）

/** * 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>();        Stack<TreeNode> nodeStack=new Stack<>();        while(true)        {            while(root!=null)            {                res.add(root.val);                nodeStack.push(root);                root=root.left;            }            if(nodeStack.isEmpty()) break;            TreeNode tempNode=nodeStack.pop();            root=tempNode.right;        }        return res;    }}

解3（非递归）

//前序遍历    非递归    public static List<TreeNode> preorderTraversal2(TreeNode root){         List<TreeNode> tList=new ArrayList<TreeNode>();         Stack<TreeNode> tStack=new Stack<TreeNode>();         tStack.push(root);         while(!tStack.isEmpty()){             TreeNode p=tStack.pop();             tList.add(p);             System.out.print(p.val+" ");             if(p.right!=null) tStack.push(p.right);             if(p.left!=null)  tStack.push(p.left);         }          return tList;    }