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?
只要是遞歸就能用棧來還原遞歸過程
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { Listres = new ArrayList<>(); Stack stack = new Stack<>(); public List preorderTraversal(TreeNode root) { if(root == null) return new ArrayList (); stack.push(root); while(!stack.isEmpty()){ TreeNode tn = stack.pop(); res.add(tn.val); if(tn.right!=null){ stack.push(tn.right); } if(tn.left!=null){ stack.push(tn.left); } } return res; } }