* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<int> postorderTraversal(TreeNode *root) { vector<int>result;if(root==NULL)returnresult; stack<TreeNode*>sTree; TreeNode*
Before you attempt this problem, you might want to try coding a pre-order traversal iterative solution first, because it is easier. On the other hand, coding a post-order iterative version is a challenge. See my post:Binary Tree Post-Order Traversal Iterative Solutionfor more details and an ...
Space: O(n). The stack would store all the tree elements in worst case. The set would store all the nodes at the end. 2.2 Iterative solution A better way is based on the fact that post-order traversal is the reverse order of pre-order traversal with visiting the right subtree before ...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: //非递归方法实现后序遍历 vector<int> postorderTraversal(TreeNode* root) { vector<int> res; if(root==NULL) return res; stack<TreeNode *> vis; stack<TreeNode *> travel; travel.push(root...
LeetCode 145 题, Binary Tree Postorder Traversal 解题思路。 1、先读题,题目要求后序遍历,用非递归解决。 递归解决很容易,代码如下: # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x ...
题目描述英文版描述Given the root of a binary tree, return the postorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [3,2,1] Example 2: Input: root = [] Output: …
Preorder traversal Inorder traversal Postorder traversal Essentially, all three traversals work in roughly the same manner. They start at the root and visit that node and its children. The difference among these three traversal methods is the order with which they visit the node itself versus visi...
Preorder traversal Inorder traversal Postorder traversal Essentially, all three traversals work in roughly the same manner. They start at the root and visit that node and its children. The difference among these three traversal methods is the order with which they visit the node itself versus visi...
binary-tree-postorder-traversal title: binary-tree-postorder-traversal 描述 Given a binary tree, return thepostordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 return[3,2,1]. Note:Recursive solution is trivial, could you do it iteratively?
* TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */classSolution{public:vector<int>postorderTraversal(TreeNode*root){vector<int>res;if(root==nullptr)returnres;stack<TreeNode*>st;st.push(root);TreeNode*pre=nullptr;while(!st.empty()){TreeNode*cur=...