145. Binary Tree Postorder Traversal

Given a binary tree, return the _postorder _traversal of its nodes' values.

For example:
Given binary tree[1,null,2,3],

   1
    \
     2
    /
   3

return[3,2,1].

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

Solution 1: Recursive, DFS O(n); O(h)

    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        dfs(res, root);
        return res;
    }

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

九章写法，不用helper函数

    public List<Integer> postorderTraversal(TreeNode root) {
        //虽然要用到ArrayList的addAll方法，但是List不用改成ArrayList
        List<Integer> result = new ArrayList<>();
        if (root == null) {
            return result;
        }

        result.addAll(postorderTraversal(root.left));
        result.addAll(postorderTraversal(root.right));
        result.add(root.val);

        return result;
    }

Solution 2: Iterative + Stack O(n); O(h)

https://leetcode.com/problems/binary-tree-postorder-traversal/discuss/45551/Preorder-Inorder-and-Postorder-Iteratively-Summarization

    public List<Integer> postorderTraversal(TreeNode root) {
        //since we need use addFirst() method, must use LinkedList
        LinkedList<Integer> res = new LinkedList<>();
        if (root == null) {
            return res;
        }
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.empty()) {
            TreeNode cur = stack.pop();
            res.addFirst(cur.val); 
            if (cur.left != null) {
                stack.push(cur.left);
            }
            if (cur.right != null) {
                stack.push(cur.right);
            }
        }
        return res;
    }

九章写法，规范不投机

   public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) {
            return result;
        }

        TreeNode prev = null; // previously traversed node
        TreeNode curr = root;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.empty()) {
            curr = stack.peek();
            if (prev == null || prev.left == curr || prev.right == curr) { // traverse down the tree
                if (curr.left != null) {
                    stack.push(curr.left);
                } else if (curr.right != null) {
                    stack.push(curr.right);
                }
            } else if (curr.left == prev) { // traverse up the tree from the left
                if (curr.right != null) {
                    stack.push(curr.right);
                }
            } else { // traverse up the tree from the right
                result.add(curr.val);
                stack.pop();
            }
            prev = curr;
        }

        return result;
    }