* @return: Postorder in ArrayList which contains node values. */vector<int>postorderTraversal(TreeNode * root){// write your code herevector<int> result; traverse(root, result);returnresult; }// 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voidtraverse(TreeNode ...
* 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...
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: …
* 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...
Binary Tree Postorder Traversal https://leetcode.com/problems/binary-tree-postorder-traversal/ Constraint: Idea: We could do it either in recursive or iterative way. Code Recursive solution: AI检测代码解析 class Solution { public List<Integer>postorderTraversal(TreeNode root) {...
A node's children are those nodes that appear immediately beneath the node itself. A node's parent is the node immediately above it. A tree's root is the single node that contains no parent.Figure 1 shows an example of the chain of command in a fictional company....
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?