145. Binary Tree Postorder Traversal

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

Example:

Input: [1,null,2,3]
   1
    
     2
    /
   3

Output: [3,2,1]

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

非递归实现二叉树的后序遍历

java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>() ;
        Stack<TreeNode> stack = new Stack<TreeNode>() ;
        stack.push(root) ;
        while(!stack.isEmpty()){
            TreeNode node = stack.pop() ;
            if (node == null)
                continue ;
            res.add(node.val) ;
            stack.push(node.left) ;
            stack.push(node.right) ;
        }
        Collections.reverse(res) ;
        return res ;
    }
}