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[Leetcode] Binary Tree Preorder Traversal
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?
Solution 1: 非递归
1 /** 2 * Definition for binary tree 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 List<Integer> result=new ArrayList<Integer>(); 13 if(root==null){14 return result;15 }16 Stack<TreeNode> s=new Stack<TreeNode>();17 result.add(root.val);18 s.add(root);19 TreeNode p=root.left;20 21 while(!s.isEmpty()){22 while(p!=null){23 s.add(p);24 result.add(p.val);25 p=p.left;26 }27 TreeNode n=s.pop();28 p=n.right;29 if(p!=null){30 s.add(p);31 result.add(p.val);32 p=p.left;33 }34 }35 return result;36 }37 }
Solution 2: 递归
1 /** 2 * Definition for binary tree 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 List<Integer> result=new ArrayList<Integer>();13 myPreorderTraversal(root,result);14 return result;15 }16 17 private void myPreorderTraversal(TreeNode root, List<Integer> result) {18 // TODO Auto-generated method stub19 if(root!=null){20 result.add(root.val);21 myPreorderTraversal(root.left, result);22 myPreorderTraversal(root.right, result);23 }24 }25 }
[Leetcode] Binary Tree Preorder Traversal
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