* 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; stack<TreeNode*>pTree; pTree.push(root);while(!pTree.empt...
* 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; stack<TreeNode*>node; s...
* Val int * Left *TreeNode * Right *TreeNode * } */funcpostorderTraversal(root*TreeNode)[]int{varresult[]intpostorder(root,&result)returnresult}funcpostorder(root*TreeNode,output*[]int){ifroot!=nil{postorder(root.Left,output)postorder(root.Right,output)*output=append(*output,root.Val)}}...
publicList<Integer>postorderTraversal(TreeNoderoot){List<Integer>list=newArrayList<>();Stack<TreeNode>stack=newStack<>();TreeNodecur=root;TreeNodelast=null;while(cur!=null||!stack.isEmpty()){if(cur!=null){stack.push(cur);cur=cur.left;}else{TreeNodetemp=stack.peek();//是否变到右子树if...
二叉树的后序遍历(Binary Tree Postorder Traversal) lintcode:题号——68,难度——easy 2 描述 给出一棵二叉树,返回其节点值的后序遍历。 名词: 遍历 按照一定的顺序对树中所有节点进行访问的过程叫做树的遍历。 后序遍历 在二叉树中,首先遍历左子树,然后遍历右子树,最后访问根结点,在遍历左右...
4. Construct Binary Tree from Inorder and Postorder Traversal 这里的解题思路是我们知道postorder的最后一个元素一定是根节点,按照这个特性可以找出根节点在inorder中的位置,从该位置分开,左右两边分别就是左右子树。 class Solution: def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNo...
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 visiting its children. To help clarify...
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 visiting its children. To help clarify...
* 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=st.top();if(((cur...
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?