145. Binary Tree Postorder Traversal
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?
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> postorderTraversal(TreeNode root) {
List<Integer> list = new LinkedList<>();
Stack<TreeNode> stack = new Stack<>();
if(root==null){
return new ArrayList<>();
}
stack.push(root);
while(!stack.isEmpty()){
TreeNode node = stack.pop();
list.add(0, node.val);
if(node.left!=null){
stack.push(node.left);
}
if(node.right!=null){
stack.push(node.right);
}
}
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> postorderTraversal(TreeNode root) {
if(root==null){
return list;
}
postOrder(root);
return list;
}
private void postOrder(TreeNode root){
if(root.left!=null){
postOrder(root.left);
}
if(root.right!=null){
postOrder(root.right);
}
list.add(root.val);
}
}