We read every piece of feedback, and take your input very seriously.
To see all available qualifiers, see our documentation.
Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.
By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.
Already on GitHub? Sign in to your account
题目描述:
给定一个二叉树,返回它的 后序 遍历。
示例:
输入: [1,null,2,3] 1 \ 2 / 3 输出: [3,2,1]
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
解题思路:使用栈后序遍历,左-右-根。其中head节点是为了记录上一个处理过的节点,因为是后序遍历,所以node会是根节点,head为其叶子节点
C++解题:
class Solution { public: vector<int> postorderTraversal(TreeNode* root) { if(!root) return {}; vector<int> result; stack<TreeNode *> ss; TreeNode* head = root; ss.push(root); while (!ss.empty()) { TreeNode* node = ss.top(); if ((!node->left && !node->right) || node->left == head || node->right == head) { result.push_back(node->val); ss.pop(); head = node; }else{ if(node->right) ss.push(node->right); if(node->left) ss.push(node->left); } } return result; } };
The text was updated successfully, but these errors were encountered:
No branches or pull requests
题目描述:
给定一个二叉树,返回它的 后序 遍历。
示例:
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
解题思路:使用栈后序遍历,左-右-根。其中head节点是为了记录上一个处理过的节点,因为是后序遍历,所以node会是根节点,head为其叶子节点
C++解题:
The text was updated successfully, but these errors were encountered: