For example:
Given binary tree{1,#,2,3},
1
\
2
/
3
return[3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
递归方法
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import
java.util.ArrayList;
public
class
Solution {
public
ArrayList<Integer> postorderTraversal(TreeNode root) {
ArrayList<Integer> list =
new
ArrayList<Integer>();
if
(root ==
null
){
return
list;
}
test(root,list);
return
list;
}
public
void
test(TreeNode node,ArrayList<Integer> list){
if
(node.left !=
null
){
test(node.left,list);
}
if
(node.right !=
null
){
test(node.right,list);
}
list.add(node.val);
}
}
非递归方法
import java.util.ArrayList;
import java.util.Stack;
public class Solution {
public ArrayList<Integer> postorderTraversal(TreeNode root) {
ArrayList<Integer> list = new ArrayList<>();
Stack<TreeNode> stack = new Stack<>();
if (root == null) return list;
stack.push(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
list.add(0, node.val);//每次插入到头部
if (node.left != null) stack.push(node.left);
if (node.right != null) stack.push(node.right);
}
return list;
}
}