Implement an iterative, pre-order traversal of a given binary tree, return the list of keys of each node in the tree as it is pre-order traversed.
Examples
5
/ \
3 8
/ \ \
1 4 11
Pre-order traversal is [5, 3, 1, 4, 8, 11]
Corner Cases
What if the given binary tree is null? Return an empty list in this case.
interative: 时间复杂度： o(n) 每个点都走， 空间复杂度 o(n) 但是是在 heap 上
1 public class Solution {
2 public List<Integer> preOrder(TreeNode root) {
3 // Write your solution here
4 List<Integer> res = new ArrayList<>();
5 if (root == null) {
6 return res ;
7 }
8 Deque<TreeNode> stack = new LinkedList<>() ;
9 stack.offerFirst(root) ;
10 while(!stack.isEmpty()){
11 TreeNode curr = stack.pollFirst() ;
12 res.add(curr.key);
13 if (curr.right != null) {
14 stack.offerFirst(curr.right);
15 }
16 if (curr.left != null) {
17 stack.offerFirst(curr.left);
18 }
19 }
20 return res ;
21 }
22 }
recursive : time: o(n) for every node, space: o(h) on the stack:
1 private List<Integer> preOrder_recur(TreeNode root){ 2 List<Integer> res = new ArrayList<>() ; 3 if (root == null) { 4 return res ; 5 } 6 helper(root, res); 7 return res ; 8 } 9 private void helper(TreeNode root , List<Integer> res ){ 10 //base case 11 if (root == null) { 12 return ; 13 } 14 res.add(root.key) ; 15 helper(root.left, res) ; 16 helper(root.right, res); 17 }