145. Binary Tree Postorder Traversal

Problem

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 (Java)

/**
 * 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) {
        if (root == null) {
            return new ArrayList<>();
        }
        List<Integer> res = postorderTraversal(root.left);
        res.addAll(postorderTraversal(root.right));
        res.add(root.val);
        return res;
    }
}

Solution 1 (Javascript)

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var postorderTraversal = function(root) {
  var res = [];
  helper(root, res);
  return res;
};

var helper = function (root, res) {
  if (!root) return;
  helper(root.left, res);
  helper(root.right, res);
  res.push(root.val);
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).

Solution 2 (Javascript)

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var postorderTraversal = function(root) {
  if (!root) return [];
  var res = [];
  var stack = [];
  var node = root;
  while (node || stack.length) {
    if (node) {
      stack.push(node);
      res.unshift(node.val);
      node = node.right;
    } else {
      node = stack.pop();
      node = node.left;
    }
  }
  return res;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).