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?
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1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public List<Integer> preorderTraversal(TreeNode root) { 12 ArrayList<Integer> anslist = new ArrayList<Integer>(); 13 if(root == null) return anslist; 14 Stack<TreeNode> s = new Stack<TreeNode>(); 15 s.push(root); 16 while(!s.empty()){ 17 TreeNode n = s.pop(); 18 anslist.add(n.val); 19 if(n.right != null) s.push(n.right); 20 if(n.left != null) s.push(n.left); 21 } 22 return anslist; 23 } 24 }
使用栈来模拟先序遍历。 递归的方法很简单,这里就不写了。