1102. Invert a Binary Tree (25)_1102. invert a binary tree (25)-pat甲级真题 1102. inve-优快云博客

本文介绍了一个关于二叉树的数据结构实现，包括如何根据输入构建二叉树、找到根节点、翻转二叉树，并进行层序和中序遍历。展示了具体的C++代码实现。

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唯一注意一点，就是根结点这里并未给出，需要自己根据输入判断出来。

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>

using namespace std;

#define INF (~(1<<31))

struct Node{
	int id, left, right;

	Node(){}
	Node(int i, int l, int r) : id(i), left(l), right(r){}
};

vector<Node> nodes;
vector<int> level;
vector<int> inorder;

void invert(int root){
	if(root == -1) return;

	swap(nodes[root].left, nodes[root].right);
	invert(nodes[root].left);
	invert(nodes[root].right);
}

void inorderTranversal(int root){
	if(root == -1) return;

	inorderTranversal(nodes[root].left);
	inorder.push_back(root);
	inorderTranversal(nodes[root].right);
}

int main(){
	int n;
	cin >> n;

	
	vector<int> father(n, -1);
	nodes.resize(n);

	for(int i = 0; i < n; ++i){
		char l, r;
		cin >> l >> r;
		int left = -1, right = -1;
		
		if(l != '-') left = l - '0';
		if(r != '-') right = r - '0';

		nodes[i] = Node(i, left, right);

		if(left != -1) father[left] = i;
		if(right != -1) father[right] = i;
	}

	int root = find_if(begin(father), end(father), [](int val){ return val == -1; }) - begin(father);

	invert(root);

	queue<int> Q;
	Q.push(root);

	while(!Q.empty()){
		int r = Q.front(); Q.pop();

		level.push_back(r);
		if(nodes[r].left != -1) Q.push(nodes[r].left);
		if(nodes[r].right != -1) Q.push(nodes[r].right);
	}

	inorderTranversal(root);

	for(size_t i = 0; i < level.size(); ++i){
		if(i) printf(" ");
		printf("%d", level[i]);
	}

	printf("\n");
	for(size_t i = 0; i < inorder.size(); ++i){
		if(i) printf(" ");
		printf("%d", inorder[i]);
	}

	return 0;
}