December 28, 2021

145. Binary Tree Postorder Traversal

Solution1: Recursion (Traverse)

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

Solution2: Recursion (Divide and Conquer)

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

Solution3: Iterative

Method1: Post-order is the reverse order of pre-order with traversing right subtree before traversing left subtree.

/**
 * 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) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        Deque<TreeNode> preOrder = new LinkedList<>(); // Stack
        preOrder.offerFirst(root);
        while (!preOrder.isEmpty()) {
            TreeNode current = preOrder.pollFirst();
            res.add(current.val);
            if (current.left != null) {
                preOrder.offerFirst(current.left);
            }
            if (current.right != null) {
                preOrder.offerFirst(current.right);
            }
        }
        Collections.reverse(res);
        return res;
    }
}

Method 2: Check the relation between the current node and the previous node to determine which direction should go next.

/**
 * 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) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        Deque<TreeNode> stack = new LinkedList<>();
        stack.offerFirst(root);
        TreeNode prev = null;
        while (!stack.isEmpty()) {
            TreeNode cur = stack.peekFirst();
            if (prev == null || cur == prev.left || cur == prev.right) {
                if (cur.left != null) {
                    stack.offerFirst(cur.left);
                } else if (cur.right != null) {
                    stack.offerFirst(cur.right);
                } else {
                    stack.pollFirst();
                    res.add(cur.val);
                }
            } else if (prev == cur.right || prev == cur.left && cur.right == null) {
                stack.pollFirst();
                res.add(cur.val);
            } else {
                stack.offerFirst(cur.right);
            }
            prev = cur;
        }
        return res;
    }
}