Total Accepted: 15614 Total Submissions: 50928 My Submissions
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?
【解題思路】
1、遞迴 左右中

Java AC

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private ArrayList<Integer> list;
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        list = new ArrayList<Integer>();
        postOrder(root);
        return list;
    }
    private void postOrder(TreeNode root){
        if(root == null){
            return;
        }
        postOrder(root.left);
        postOrder(root.right);
        list.add(root.val);
    }
}
 

2、非遞迴
1)、需要先查詢到當前節點的最左孩子，也就是一直入棧。
2)、假設到node時，左孩子為空了，這個時候出棧，棧頂元素就是當前節點node。
3)、判斷棧頂元素是否為空，如果為空，說明可以直接訪問根節點了，直接加入list中。
4)、如果不為空，說明需要訪問右孩子，這個時候，注意將當前節點元素new一個出來，同時入棧，這一步其實是儲存根節點值的。再回到1)重複處理，直到棧為空。

Java AC

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private ArrayList<Integer> list;
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        list = new ArrayList<Integer>();
        postOrder(root);
        return list;
    }
    private void postOrder(TreeNode root){
        Stack<TreeNode> stack = new Stack<TreeNode>();
        while(root != null || !stack.isEmpty()){
            while(root != null){
                stack.push(root);
                root = root.left;
            }
            while(!stack.isEmpty()){
                root = stack.pop();
                if(root.right != null){
                	stack.push(new TreeNode(root.val));
                    root = root.right;
                    break;
                }
                list.add(root.val);
                if (stack.isEmpty()) {
					return;
				}
            }
        }
    }
}