* 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*pCur=root; TreeNode*pPre=NULL;while(!sTree.empty()||pCur) {while(...
Given a binary tree, return the postorder traversal 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?We know the elements can be printed post-order easily using recursion, as ...
* 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...
publicList<Integer>postorderTraversal(TreeNoderoot){List<Integer>list=newArrayList<>();if(root==null){returnlist;}Stack<TreeNode>stack=newStack<>();stack.push(root);stack.push(root);while(!stack.isEmpty()){TreeNodecur=stack.pop();if(cur==null){continue;}if(!stack.isEmpty()&&cur==stac...
145. Binary Tree Postorder Traversal 二叉树的后序遍历 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。
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...
1. Problem Descriptions:Given two integer arrays inorderandpostorderwhereinorderis the inorder traversal of a binary tree andpostorderis the postorder traversal of the same tree, construct and retu…
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...
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=...