LeetCode 145. Binary Tree Postorder Traversal--后序遍历--先序遍历反向输出--递归，迭代--C++,Python解法

题目地址：Binary Tree Postorder Traversal - LeetCode

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

Example:

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

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

经典的二叉树后续遍历，最简单的递归做法。
Python解法如下：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def postorderTraversal(self, root: TreeNode) -> List[int]:def helper(root, l):if root is None:returnhelper(root.left, l)helper(root.right, l)l.append(root.val)l = []helper(root, l)return l

迭代做法要麻烦一点，不容易想到。
比较好的做法是先序遍历的结果反向输出。
Python解法如下：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def postorderTraversal(self, root: TreeNode) -> List[int]:if root is None:return []stack, output = [root, ], []while stack:root = stack.pop()output.append(root.val)if root.left is not None:stack.append(root.left)if root.right is not None:stack.append(root.right)return output[::-1]

时间复杂度为O(n)，空间复杂度为O(1)。

C++解法如下：

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {public:vector<int> postorderTraversal(TreeNode *root) {vector<int> result;stack<TreeNode *> l;if (root == nullptr) {return result;}l.push(root);while (!l.empty()) {auto curr = l.top();l.pop();result.push_back(curr->val);if (curr->left != nullptr) {l.push(curr->left);}if (curr->right != nullptr) {l.push(curr->right);}}std::reverse(result.begin(), result.end());return result;}
};

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