leetcode 145. Binary Tree Postorder Traversal-后续遍历|递归|非递归

原题链接：145. Binary Tree Postorder Traversal
【思路-Java、Python】-非递归实现

以[1,2,3,4,5,null,6]为例，需要用到一个栈，具体执行过程如下图所示：



stack 取出的节点值是从 res 的头部插入的，这点与二叉树的前序遍历、中序遍历存在不同：

public class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
Stack<TreeNode> stack = new Stack<TreeNode>();
List<Integer> res = new ArrayList<Integer>();
stack.add(root);
while (!stack.isEmpty()) {
TreeNode temp = stack.pop();
if (temp != null) {
res.add(0, temp.val);
stack.add(temp.left);
stack.add(temp.right);
}
}
return res;
}
}67 / 67 test
cases passed. Runtime: 2
ms  Your runtime beats 7.50% of javasubmissions.

class Solution(object):
def postorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
stack = []
stack.append(root)
res = []
while stack :
temp = stack.pop()
if temp :
res.append(temp.val)
stack.append(temp.left)
stack.append(temp.right)
return res[::-1]67 / 67 test
cases passed. Runtime: 40
ms  Your runtime beats 67.45% of pythonsubmissions.

【补充-Python】-递归实现

递归方法很简单，一般都能想到，这里也就不赘述了，用 Python 的递归实现

class Solution(object):
def postorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
res = []
def dfs(root, res) :
if root :
dfs(root.left, res)
dfs(root.right, res)
res.append(root.val)
dfs(root, res)
return res67 / 67 test
cases passed. Runtime: 40
ms  Your runtime beats 67.45% of pythonsubmissions.