589. N-ary Tree Preorder Traversal
Given an n-ary tree, return the preorder traversal of its nodes' values.
For example, given a 3-ary tree:
Return its preorder traversal as: [1,3,5,6,2,4].
Note: Recursive solution is trivial, could you do it iteratively?
Solution (Iteration)
Using stack, push the child from the end of list
class Solution {
public List<Integer> preorder(Node root) {
List<Integer> res = new ArrayList<>();
if (root == null) return res;
Stack<Node> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
Node cur = stack.pop();
res.add(cur.val);
for (int i = cur.children.size()-1; i >= 0; i--) stack.push(cur.children.get(i));
}
return res;
}
}
Solution (Recursion)
/*
// Definition for a Node.
class Node {
public int val;
public List<Node> children;
public Node() {}
public Node(int _val,List<Node> _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public List<Integer> preorder(Node root) {
List<Integer> res = new ArrayList<>();
helper(root, res);
return res;
}
private void helper(Node root, List<Integer> res) {
if (root == null) return;
res.add(root.val);
for (Node node: root.children) {
helper(node, res);
}
}
}
144. Binary Tree Preorder Traversal
Given a binary tree, return the preorder traversal of its nodes' values.
Example:
Input: [1,null,2,3]
1
\
2
/
3
Output: [1,2,3]
Follow up: Recursive solution is trivial, could you do it iteratively?
Solution (Iteration)
Use stack, first push node.right, then push node.left
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
if (root == null) return res;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode cur = stack.pop();
res.add(cur.val);
if (cur.right != null) stack.push(cur.right);
if (cur.left != null) stack.push(cur.left);
}
return res;
}
}
Solution (Recursion)
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
helper(root, res);
return res;
}
private void helper(TreeNode root, List<Integer> res) {
if (root == null) return;
res.add(root.val);
helper(root.left, res);
helper(root.right, res);
}
}
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