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144. Binary Tree Preorder Traversal

求阙者 2022-08-03 阅读 43


Given a binary tree, return the preorder traversal of its nodes’ values.

For example:
Given binary tree [1,null,2,3],

   1
\
2
/
3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> list = new ArrayList<Integer>();

public List<Integer> preorderTraversal(TreeNode root) {
preOrder(root);
return list;
}

public void preOrder(TreeNode node) {
if (node == null) {
return;
}
list.add(node.val);
if (node.left != null) {
preOrder(node.left);
}
if (node.right != null) {
preOrder(node.right);
}
}
}

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
Stack<TreeNode> stack = new Stack<>();
TreeNode head = root;
while(head != null || stack.size() > 0){
if(head != null){
res.add(head.val);
if(head.right != null){
stack.push(head.right);
}
head = head.left;
}else{
head = stack.pop();
}
}
return res;
}
}


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