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?
Binary Tree
的问题,递归法是最简单的。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
if (root == null) { return res; }
res.add(root.val);
res.addAll(preorderTraversal(root.left));
res.addAll(preorderTraversal(root.right));
return res;
}
}
用一个Stack
缓存接下来需要处理的节点。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
Deque<TreeNode> stack = new LinkedList<TreeNode>();
stack.add(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pollFirst();
if (node != null) {
res.add(node.val);
stack.offerFirst(node.right);
stack.offerFirst(node.left);
}
}
return res;
}
}