binary-tree-preorder-traversal

题目来源：leetcode

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].

思路：dfs递归（题目简单不要脸）

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */import java.util.*;public class Solution {    public ArrayList<Integer> preorderTraversal(TreeNode root) {        ArrayList<Integer> res=new ArrayList<>();        if (root==null)            return res;        dfs(res,root);        return res;    }    public void dfs(ArrayList<Integer> ls,TreeNode root){        ls.add(root.val);        if (root.left!=null)            dfs(ls,root.left);        if (root.right!=null)            dfs(ls,root.right);    }}

中序遍历

/** * 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> inorderTraversal(TreeNode root) {        List<Integer> ls=new ArrayList<Integer>();        if(root==null)            return ls;        dfs(ls,root);        return ls;    }    public void dfs(List<Integer> ls,TreeNode root){        if(root.left!=null)            dfs(ls,root.left);            ls.add(root.val);        if(root.right!=null)            dfs(ls,root.right);    }}