* Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<int> postorderTraversal(TreeNode *root) { vector<int>result;if(root==NULL)returnresult...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ //中介节点 structTempNode { TreeNode *btnode; boolisFirst; }; //非递归后序遍历-迭代 vector<int> postorderTraversal(TreeNode *root) { stack<TempNode *> s; vector<int> path; TreeNode *p = root; TempNode *tem...
vector<int> postorderTraversal(TreeNode * root) { // write your code here vector<int> result; traverse(root, result); return result; } // 遍历法的递归函数,没有返回值,函数参数result贯穿整个递归过程用来记录遍历的结果 void traverse(TreeNode * root, vector<int> & result) // 递归三要素之定...
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] 1. 2. 3. 4. 5. 6. 7. 8. 进阶:递归算法很简单,你可以通过迭代算法完成吗? DFS 今天这道题比较简单,直接DFS就可以了。 Code def postorderTraversal(self, root: TreeNode) -> List[int]: def dfs...
publicList<Integer>inorderTraversal(TreeNoderoot){List<Integer>ans=newArrayList<>();Stack<TreeNode>stack=newStack<>();TreeNodecur=root;while(cur!=null||!stack.isEmpty()){//节点不为空一直压栈while(cur!=null){stack.push(cur);cur=cur.left;//考虑左子树}//节点为空,就出栈cur=stack.pop()...
The function returns the root of the constructed binary tree. 4. Time & Space Complexity Analysis: 4.1 Time Complexity: 4.1.1 Lists as parameters In each recursive call, the index() function is used to find the index of the root value in the inorder traversal list. This function has a ...
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?
publicvector<int>postorderTraversal(TreeNode*root){vector<int>result;stack<TreeNode*>st;TreeNode*p=root,*q=NULL;do{while(p)st.push(p),p=p->left;q=NULL;while(!st.empty()){p=st.top();st.pop();if(p->right==q)//右子树为空或者已经访问{result.push_back(p->val);q=p;}else//...
Binary tree traversal: Preorder, Inorder, and Postorder In order to illustrate few of the binary tree traversals, let us consider the below binary tree: Preorder traversal: To traverse a binary tree in Preorder, following operations are carried-out (i) Visit the root, (ii) Traverse the le...
Can you solve this real interview question? Binary Tree Postorder Traversal - 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] Explanation: [https://assets