入栈顺序为先序遍历(根左右),出栈顺序为中序遍历(左根右)。 1086 Tree Traversals Again #include<bits/stdc++.h>usingnamespacestd;typedeflonglongll;constintN=1e6+100;constintM=N*10;constintmod=1e9+7;constdoubleeps=1e-9;typedefpair<int,int>PII;#define endl '\n'#define x first#define y ...
1086. Tree Traversals Again (25) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered...
Node* lchild; Node* rchild; }; stack<int> s;intin[maxn], pre[maxn];intn, num =0;voidPostOrder(Node* root){if(root ==NULL){return; }PostOrder(root->lchild);PostOrder(root->rchild);printf("%d", root->data);if(num < n-1){printf(" "); num++; } }Node*create(intpreL,intp...
输入一个正整数N(<=30),接着输入2*N行表示栈的出入(入栈顺序表示了二叉搜索树的先序序列,出栈顺序表示了二叉搜索树的中序序列),输出后序序列。 AAAAAccepted code: 1#defineHAVE_STRUCT_TIMESPEC2#include<bits/stdc++.h>3usingnamespacestd;4stack<int>st;5inta[37],b[37];6intans[37];7voidbuild(in...
简介:【1086】Tree Traversals Again (25 分)【1086】Tree Traversals Again (25 分) #include<iostream>#include<stdio.h>#include<stdlib.h>#include<math.h>#include<string.h>#include<algorithm>#include#include<vector>#include<queue>#include<stack>using namespace std;const int maxn=50;struct node...
B1086 Tree Traversals Again (25分) 由题意可以推出,每次访问一个新的结点时就把它入栈,这个过程和先序序列总是先访问根节点的性质是相同的,因此Push的次序就是先序遍历中元素的顺序,Pop则是按左子树,根节点,右子树的顺序进行的,因此Pop的次序是中序遍历序列中元素的顺序. ...
For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line. ...
【摘要】 1086 Tree Traversals Again (25 分) 题目传送门:1086 Tree Traversals Again (25 分) 一、题目大意 这道题的难点就在于是否能读懂题意了。... 1086 Tree Traversals Again (25 分) 题目传送门:1086 Tree Traversals Again (25 分) 一、题目大意 ...
1086 Tree Traversals Again 题目 题意:给出前序和中序遍历的结果,重构二叉树并返回后序遍历的结果 tip:前序中序构造后序 #include<iostream> #include<stack> #include<vector> using namespace std; vector<int>pre,inorder,pos; int n,count;
这道题我以为是考验如何运用栈来实现树的遍历。代码中有两个细节,一是要注意区分树的“NULL”状态和存在状态,二是要注意出栈和入栈的前后对称,每一次操作的根节点都是上一次出栈/入栈的value。 #include<iostream>#include<stdio.h>#include<vector>#include<stack>#include<string>usingnamespacestd;structNode{...