* 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; stack<TreeNode*>pTree; pTree.push(root);while...
classSolution {public: vector<int> postorderTraversal(TreeNode *root) { vector<int>res;if(root==NULL)returnres; stack<pair<TreeNode*,int>>s;intunUsed; s.push(make_pair(root,1));while(!s.empty()) { root=s.top().first; unUsed=s.top().second; s.pop();if(unUsed){ s.push(make...
* 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>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()...
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...
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…
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...
给你一棵二叉树的根节点root,返回其节点值的后序遍历。 示例1: 输入:root = [1,null,2,3] 输出:[3,2,1] 解释: 示例2: 输入:root = [1,2,3,4,5,null,8,null,null,6,7,9] 输出:[4,6,7,5,2,9,8,3,1] 解释: 示例3: 输入:root = [] ...