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?


Recursive DFS Solution

/**
 * 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;
 *     }
 * }
 */
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> ret = new ArrayList<>();
        dfs(root, ret);
        return ret;
    }

    private void dfs(TreeNode root, List<Integer> ret) {
        if (root != null) {
            dfs(root.left, ret);
            dfs(root.right, ret);
            ret.add(root.val);
        }
    }
}

Iterative DFS Solution

/**
 * 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) {
        LinkedList<Integer> ans = new LinkedList<>();
        Stack<TreeNode> stack = new Stack<>();
        if (root == null) return ans;

        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            ans.addFirst(cur.val);
            if (cur.left != null) {
                stack.push(cur.left);
            }
            if (cur.right != null) {
                stack.push(cur.right);
            } 
        }
        return ans;
    }
}

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