[Leetcode] Binary Tree Postorder Traversal

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?

Solution 1: 非递归

/**
 * 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) {
        List<Integer> result = new ArrayList<Integer>();
        if (root == null)
            return result;
        Stack<TreeNode> s = new Stack<TreeNode>();
        TreeNode node = root, prev = root;
        while (node != null || !s.isEmpty()) {
            while (node != null) {
                s.add(node);
                node = node.left;
            }
            if (s.size() > 0) {
                TreeNode temp = s.peek().right;
                if(temp!=null&&temp!=prev){
                    node=temp;
                }else if(temp==prev||temp==null){
                    node=s.pop();
                    result.add(node.val);
                    prev=node;
                    node=null;
                }
            }
        }
        return result;
    }
}

 

Solution 2: 递归

/**
 * 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) {
        List<Integer> result=new ArrayList<Integer>();
        myPostorderTraversal(root,result);
        return result;
    }

    private void myPostorderTraversal(TreeNode root, List<Integer> result) {
        // TODO Auto-generated method stub
        if(root!=null){
            myPostorderTraversal(root.left, result);
            myPostorderTraversal(root.right, result);
            result.add(root.val);
        }
    }
}