给你一棵二叉树的根节点 root ，返回其节点值的 后序遍历 。
示例 1：
输入： root = [1,null,2,3]
输出： [3,2,1]
解释：
示例 2：
输入： root = [1,2,3,4,5,null,8,null,null,6,7,9]
输出： [4,6,7,5,2,9,8,3,1]
解释：
示例 3：
输入： root = []
输出： []
示例 4：
输入： root = [1]
输出： [1]
提示：
- 树中节点的数目在范围
[0, 100]内 -100 <= Node.val <= 100
进阶： 递归算法很简单，你可以通过迭代算法完成吗？
解题答案
/**
* 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> postorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
if (root == null) return res;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
TreeNode lastPopNode = root;
while (!stack.isEmpty()) {
TreeNode topNode = stack.peek();
if ((topNode.left == null && topNode.right == null) ||
(topNode.left == lastPopNode || topNode.right == lastPopNode)) {
lastPopNode = stack.pop();
res.add(lastPopNode.val);
} else {
if (topNode.right != null) {
stack.push(topNode.right);
}
if (topNode.left != null) {
stack.push(topNode.left);
}
}
}
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 {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
postorder(root, res);
return res;
}
private void postorder(TreeNode node, List<Integer> res) {
if (node == null) return;
postorder(node.left, res);
postorder(node.right, res);
res.add(node.val);
}
}
