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1127 ZigZagging on a Tree (30 分)
蛇形输出层序遍历，这个只要记录每一层的遍历序列，然后输出之前该翻转的翻转一下就可以了，然后看了看自己一年之前提交的，思路真的是一模一样，就是代码风格变了， #include <bits/stdc++.h> #define fi first #define se second #define pb push_back #define all(x) (x).begin(), (PAT 1127 ZigZagging on a Tree
题解： 代码： #include<iostream> #include<vector> using namespace std; const int maxv=39; vector<int> tree[maxv]; int in[maxv],post[maxv]; int layer=1; void layertravel(int postL,int postR,int inL,int inR,int nowlayer){ if(layer<nowlaye1127 ZigZagging on a Tree（30分）
假设二叉树中的所有键都是不同的正整数。可以通过给定的一对后序遍历和有序遍历序列来确定唯一的二叉树。这是一个简单的标准例程，可以按级别顺序打印数字。但是，如果您认为问题太简单了，那就太天真了。这次，您应该按“曲折顺序”打印数字-也就是说，从根开始，逐级打印数字，从左到右和从右1127 ZigZagging on a Tree (30分)
Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. HowevePAT 1127 ZigZagging on a Tree(30分)
Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. Howeve1127 ZigZagging on a Tree (30 分)
1127 ZigZagging on a Tree (30 分) Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to prin1127 ZigZagging on a Tree (30 分)
1127 ZigZagging on a Tree (30 分) Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routinePAT A1127 ZigZagging on a Tree （30 分）
Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. Howeve