题目描述
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?
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ import java.util.*; public class Solution { ArrayList<Integer> list = new ArrayList<Integer>(); public ArrayList<Integer> preorderTraversal(TreeNode root) { if(root == null) return list; Stack<TreeNode> stack = new Stack<TreeNode>(); stack.push(root); while(!stack.isEmpty()) { TreeNode te = stack.pop(); list.add(te.val); if(te.right != null) { stack.push(te.right); } if(te.left != null) { stack.push(te.left); } } return list; } }
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ //递归,先序遍历:根 左 右 import java.util.*; public class Solution { public ArrayList<Integer> preorderTraversal(TreeNode root) { ArrayList<Integer> list = new ArrayList<Integer>(); if(root == null){ //判断root==null return list; //{}返回为[] } list.add(root.val); list.addAll(preorderTraversal(root.left)); //addAll:添加集合 list.addAll(preorderTraversal(root.right)); return list; } }