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...
代码如下: 1#Definition for a binary tree node.2#class TreeNode(object):3#def __init__(self, x):4#self.val = x5#self.left = None6#self.right = None78classSolution(object):9defpostorderTraversal(self, root):10"""11:type root: TreeNode12:rtype: List[int]13"""14res =[]15tmp ...
* 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...
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? 本题难度Hard。有3种算法分别是: 递归...
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…
publicList<Integer>postorderTraversal(TreeNoderoot){List<Integer>list=newArrayList<>();Stack<TreeNode>stack=newStack<>();Set<TreeNode>set=newHashSet<>();TreeNodecur=root;while(cur!=null||!stack.isEmpty()){while(cur!=null&&!set.contains(cur)){stack.push(cur);cur=cur.left;}cur=stack....
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...
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? 这道题想了很久,后序遍历应该是最麻烦的吧,最容易想到的方法就是直接设置标志位...