LeetCode-in-Java
145. Binary Tree Postorder Traversal
Easy
Given the root of a binary tree, return the postorder traversal of its nodes’ values.
Example 1:
Input: root = [1,null,2,3]
Output: [3,2,1]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Example 4:
Input: root = [1,2]
Output: [2,1]
Example 5:
Input: root = [1,null,2]
Output: [2,1]
Constraints:
- The number of the 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?
Solution
import com_github_leetcode.TreeNode;
import java.util.ArrayList;
import java.util.List;
/*
* 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;
* }
* }
*/
public class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
if (root == null) {
return new ArrayList<>();
}
List<Integer> res = postorderTraversal(root.left);
res.addAll(postorderTraversal(root.right));
res.add(root.val);
return res;
}
}