Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [3,2,1].

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

遞迴：

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

非遞迴：後序遍歷左右中，考慮到棧的先進先出，故先將右子樹壓棧，兩種情況記錄答案出棧，1是遇到葉子，2是父節點的子節點已經被訪問過，根據壓棧順序，可以保證這樣可以完成後序遍歷

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        Stack<TreeNode> stk = new Stack<>();
        if(root != null) {
            stk.push(root);
        }
        TreeNode cur = null;
        TreeNode pre = null;
        while(!stk.empty()) {
            cur = stk.peek();
            if(cur.left == null && cur.right == null) {
                ans.add(cur.val);
                pre = cur;
                stk.pop();
            }
            else if(pre != null && (cur.left == pre || cur.right == pre)) {
                ans.add(cur.val);
                pre = cur;
                stk.pop();
            }
            else {
                if(cur.right != null) {
                    stk.push(cur.right);
                }
                if(cur.left != null) {
                    stk.push(cur.left);
                }
            }
        }
        return ans;
    }
}