144. Binary Tree Preorder Traversal
Given the root of a binary tree, return the preorder traversal of its nodes’ values.
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| Example 1:
Input: root = [1,null,2,3]
Output: [1,2,3]
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| Example 2:
Input: root = []
Output: []
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| Example 3:
Input: root = [1]
Output: [1]
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| Example 4:
Input: root = [1,2]
Output: [1,2]
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| Example 5:
Input: root = [1,null,2]
Output: [1,2]
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Constraints:
- The number of nodes in the tree is in the range [0, 100].
- -100 <= Node.val <= 100
Follow up: Recursive solution is trivial, could you do it iteratively?
Jamboard#
Solution#
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| /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode node) {
List<Integer> res = new ArrayList<>();
if (node == null) return res;
TreeNode[] q = new TreeNode[1000];
int n = 0;
q[n++] = node;
while (n > 0) {
TreeNode t = q[--n];
res.add(t.val);
if (t.right != null) q[n++] = t.right;
// add left as last element to retrive it first from the q
if (t.left != null) q[n++] = t.left;
}
return res;
}
public List<Integer> preorderTraversal2(TreeNode node) {
List<Integer> res = new ArrayList<>();
visit(node, res);
return res;
}
void visit(TreeNode node, List<Integer> res) {
if (node == null) return;
res.add(node.val);
visit(node.left, res);
visit(node.right, res);
}
}
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