Postorder Traversal
Given a binary tree, return the postorder traversal of its nodes' values.
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?
Solution
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def postorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if root is None:
return []
stack, result = [], []
stack.append(root)
while stack:
node = stack.pop()
result.append(node.val)
if node.left is not None: stack.append(node.left)
if node.right is not None: stack.append(node.right)
result.reverse()
return result
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
LinkedList<Integer> result = new LinkedList<>();
Stack<TreeNode> stack = new Stack<>();
if(root == null)
return result;
stack.push(root);
while(!stack.isEmpty()) {
TreeNode cur = stack.pop();
result.addFirst(cur.val);
if(cur.left != null)
stack.push(cur.left);
if(cur.right != null)
stack.push(cur.right);
}
return result;
}
}