给定一个二叉树,返回它的 后序 遍历。
示例:
输入: [1,null,2,3] 1 \ 2 / 3 输出: [3,2,1]
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
递归遍历。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
def postorder(root):
if root:
postorder(root.left)
postorder(root.right)
res.append(root.val)
res = []
postorder(root)
return res
/**
* 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> res;
public List<Integer> postorderTraversal(TreeNode root) {
res = new ArrayList<>();
postorder(root);
return res;
}
private void postorder(TreeNode root) {
if (root != null) {
postorder(root.left);
postorder(root.right);
res.add(root.val);
}
}
}