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

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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?

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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 = None

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        def helper(root, l):
            if root is None:
                return
            helper(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 = None

class 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;
    }
};