题目:
Given a binary tree, return the preorder traversal of its nodes' values
Note: Recursive solution is trivial, could you do it iteratively?
给定二叉树,返回节点值的前序遍历
注意:递归的解决方法很简单,你能用迭代地这样做吗?
思路:
还是递归比较好用,但是题目让用迭代,可以利用栈,但要注意先压入右孩子
说下add和addAll区别:
result.addAll(list);//把list中的每一个元素加到result中,result.size()==list.size()
result.add(list);//将list作为一个元素加到result中,则result.size()为1
先贴个递归的代码:
import java.util.*;
public class Solution {
public ArrayList<Integer> preorderTraversal(TreeNode root) {
ArrayList<Integer> arr = new ArrayList<>();
if(root == null)
return arr;
arr.add(root.val);
arr.addAll(preorderTraversal(root.left));
arr.addAll(preorderTraversal(root.right));
return arr;
}
}
迭代:
import java.util.*;
public class Solution {
public ArrayList<Integer> preorderTraversal(TreeNode root) {
ArrayList<Integer> arr = new ArrayList<>();
if(root == null)
return arr;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
TreeNode temp = null;
while(!stack.isEmpty()){
temp = stack.pop();
arr.add(temp.val);
if(temp.right != null)
stack.push(temp.right);
if(temp.left != null)
stack.push(temp.left);
}
return arr;
}
}
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