Binary Tree Postorder Traversal

Given a binary tree, return the_postorder_traversal of its nodes' values.

Example:

Input:
[1,null,2,3]

   1
    \
     2
    /
   3


Output:
[3,2,1]

Follow up:Recursive solution is trivial, could you do it iteratively?

Solution

Recursive

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        recursivePreorderTraversal(root, result);
        return result;
    }
    void recursivePreorderTraversal(TreeNode root, List<Integer> result) {
        if (root != null) {
            recursivePreorderTraversal(root.left, result);
            recursivePreorderTraversal(root.right, result);
            result.add(root.val);
        }
    }
}

LeetCode Official Solution

class Solution {
  public List<Integer> postorderTraversal(TreeNode root) {
    LinkedList<TreeNode> stack = new LinkedList<>();
    LinkedList<Integer> output = new LinkedList<>();
    if (root == null) {
      return output;
    }

    stack.add(root);
    while (!stack.isEmpty()) {
      TreeNode node = stack.pollLast();
      output.addFirst(node.val);
      if (node.left != null) {
        stack.add(node.left);
      }
      if (node.right != null) {
        stack.add(node.right);
      }
    }
    return output;
  }
}

Iterative - Using Stack

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        LinkedList<Integer> result = new LinkedList<>();
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode p = root;
        while(!stack.isEmpty() || p != null) {
            if(p != null) {
                stack.push(p);
                result.addFirst(p.val);  // Reverse the process of preorder
                p = p.right;             // Reverse the process of preorder
            } else {
                TreeNode node = stack.pop();
                p = node.left;           // Reverse the process of preorder
            }
        }
        return result;
    }
}