7-3 Left-View of Binary Tree (25分)

The left-view of a binary tree is a list of nodes obtained by looking at the tree from left hand side and from top down. For example, given a tree shown by the figure, its left-view is { 1, 2, 3, 4, 5 }

Given the inorder and preorder traversal sequences of a binary tree, you are supposed to output its left-view.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤20), which is the total number of nodes in the tree. Then given in the following 2 lines are the inorder and preorder traversal sequences of the tree, respectively. All the keys in the tree are distinct positive integers in the range of int.

Output Specification:

For each case, print in a line the left-view of the tree. All the numbers in a line are separated by exactly 1 space, and there must be no extra space at the beginning or the end of the line.

Sample Input:

2 3 1 5 4 7 8 61 2 3 6 7 4 5 8

Sample Output:

1 2 3 4 5

题目限度：

题目粗心：

给定一颗二叉树的先序和中序序列，须要输入该二叉树的左视图，也就是每一层最右边的结点。

算法思路：

先依据先序和中序进行建树，而后应用档次遍历获取每一个结点的根节点，并应用currentLevel记录以后结点所处档次，在结点的档次第一次发生变化的时候就是每一层的最左结点，而后应用result数组进行保留，并更新以后节点所处档次，最初输入即可

提交后果：

AC代码：

#include<cstdio>#include<queue>#include<unordered_map>using namespace std;struct Node{    int data;    Node* left;    Node* right;    int level;};int N;//节点个数int pre[30],in[30];unordered_map<int,int> pos;//每一个节点在中序序列中的个数Node* createTree(int preL,int preR,int inL,int inR){    if(preL>preR) return nullptr;    Node* root = new Node;    root->data = pre[preL];    int k = pos[root->data];// 根节点在中序中的地位    int numOfLeft = k-inL;    root->left = createTree(preL+1,preL+numOfLeft,inL,k-1);    root->right = createTree(preL+numOfLeft+1,preR,k+1,inR);    return root;}int currentLevel = 0;vector<int> result;void BFS(Node* root){    root->level = 1;    queue<Node*> q;    q.push(root);    while (!q.empty()){        Node* t = q.front();        q.pop();        if(currentLevel!=t->level){            // 达到节点档次转折处            result.push_back(t->data);            currentLevel = t->level;        }        if(t->left){            t->left->level = t->level+1;            q.push(t->left);        }        if(t->right){            t->right->level = t->level+1;            q.push(t->right);        }    }}int main(){    scanf("%d",&N);    for (int i = 0; i < N; ++i) {        scanf("%d",&in[i]);        pos[in[i]] = i;    }    for(int i=0;i<N;++i){        scanf("%d",&pre[i]);    }    Node* root = createTree(0,N-1,0,N-1);    BFS(root);    for(int i=0;i<result.size();++i){        printf("%d",result[i]);        if(i<result.size()-1) printf(" ");    }    return 0;}