144. Binary Tree Preorder Traversal
Given the root of a binary tree, return the preorder traversal of its nodes' values.
Example 1:
Input: root = [1,null,2,3] Output: [1,2,3] Example 2:
Input: root = [] Output: []
Example 3:
Input: root = [1] Output: [1]
Example 4:
Input: root = [1,2] Output: [1,2]
Example 5:
Input: root = [1,null,2] Output: [1,2]
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?
非递归
/**
* 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 root) {
Stack<TreeNode> stack = new Stack<>();
List<Integer> list = new ArrayList<>();
if (root == null){
return list;
}
stack.push(root);
while(!stack.isEmpty()){
TreeNode temp = stack.pop();
list.add(temp.val);
if(temp.right!=null){
stack.add(temp.right);
}
if(temp.left!=null){
stack.add(temp.left);
}
}
return list;
}
}递归
/**
* 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 {
private List<Integer> list = new ArrayList<>();
public List<Integer> preorderTraversal(TreeNode root) {
if(root==null){
return list;
}
preOrder(root);
return list;
}
public void preOrder(TreeNode root){
list.add(root.val);
if(root.left!=null){
preOrder(root.left);
}
if(root.right!=null){
preOrder(root.right);
}
}
}