使用迭代法对二叉树进行后序遍历——Leetcode系列（六）

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?

My Answer

<span style="font-size:14px;">/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        List<Integer> list = new ArrayList<Integer>();
        if(root == null){
            return list;
        }
        stack.push(root);
        while(!stack.empty()){
            TreeNode top = stack.pop();
            if(top.left == null && top.right == null){
                list.add(top.val);
            }else{
                stack.push(new TreeNode(top.val));
                if(top.right != null){
                    stack.push(top.right);
                }
                if(top.left != null){
                    stack.push(top.left);
                }
            }
        }
        return list;
    }
}</span>

题目来源：https://oj.leetcode.com/problems/binary-tree-postorder-traversal/