开始看了中序遍历觉得后序遍历的非递归方法也会很容易，而后自己开始尝试编写程序，发现并不是这样的后序遍历的实现很复杂
直接放代码
Java版的递归调用
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/*
* @param root: A Tree
* @return: Postorder in ArrayList which contains node values.
*/
public List<Integer> postorderTraversal(TreeNode root) {
LinkedList<Integer> targetList = new LinkedList<>();
if(root == null)
return null;
if(root.left!=null)
targetList.addAll(postorderTraversal(root.left));
if(root.right!=null)
targetList.addAll(postorderTraversal(root.right));
targetList.add(root.val);
return targetList;
}
}
递归版的非常的容易，在此不在讲解
下面放上非递归版的代码
Python版非递归：
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
"""
@param: root: A Tree
@return: Postorder in ArrayList which contains node values.
"""
def postorderTraversal(self, root):
if root is None:
return []
temp = [] #用栈结构存储节点
target = [] #结果列表
last = None #用来存储最后的节点
cur = root
while cur is not None: #循环遍历到左子树底
temp.append(cur)
cur = cur.left
while len(temp) > 0:
cur = temp.pop() #出栈
if cur.right is None or last is cur.right: #这里进行判断，若当前的根节点没有右子树，或者当前的节点的右节点已经被标注成使用过的节点
target.append(cur.val)
last = cur #将当前的节点标注成使用过的节点
else:
temp.append(cur) #如果不是则将根节点放回temp中继续遍历右子树
cur = cur.right
while cur is not None:#右子树继续遍历左子树
temp.append(cur)#入栈
cur = cur.left
return target
每次一定要去努力思考再去借鉴别人的思路！！！！！！！！
