PAT(Advanced)1102 Invert a Binary Tree C++实现

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PAT(Advanced)甲级1102 Invert a Binary Tree C++实现

题目链接

1102 Invert a Binary Tree

题目大意

给定二叉树将其左右子树逆转，输出逆转后层次遍历和中序遍历的序列

算法思路

根据题意及二叉树先序遍历定义，找出根节点，并从根节点开始，将二叉树根节点的左右子树交换，再递归地转换根节点的左右子树，根节点子树的左右子树也必须交换，递归地，所有结点的左右子树都交换一次即可，在输出层次遍历序列和中序遍历序列

void invert(int p) {
    if (p == -1) {
        return;
    }
    swap(binaryTree[p].left, binaryTree[p].right);
    invert(binaryTree[p].left);
    invert(binaryTree[p].right);
}

AC代码

/*
author : eclipse
email  : eclipsecs@qq.com
time   : Fri Jun 26 23:23:37 2020
*/
#include <bits/stdc++.h>
using namespace std;

struct Node {
    int left;
    int right;
};

bool flag = true;

vector<Node> binaryTree;
int root;

void levelOrderTraverse() {
    queue<int> q;
    q.push(root);
    while (!q.empty()) {
        int front = q.front();
        q.pop();
        if (binaryTree[front].left != -1) {
            q.push(binaryTree[front].left);
        }
        if (binaryTree[front].right != -1) {
            q.push(binaryTree[front].right);
        }
        if (front == root) {
            printf("%d", front);
        } else {
            printf(" %d", front);
        }
    }
}

void inOrderTraverse(int p) {
    if (p == -1) {
        return;
    }
    inOrderTraverse(binaryTree[p].left);
    if (flag) {
        printf("%d", p);
        flag = false;
    } else {
        printf(" %d", p);
    }
    inOrderTraverse(binaryTree[p].right);
}

void invert(int p) {
    if (p == -1) {
        return;
    }
    swap(binaryTree[p].left, binaryTree[p].right);
    invert(binaryTree[p].left);
    invert(binaryTree[p].right);
}

int main(int argc, char const *argv[]) {
    int N;
    scanf("%d", &N);
    binaryTree.resize(N);
    vector<bool> children;
    children.resize(N);
    for (int i = 0; i < N; i++) {
        string left, right;
        cin >> left >> right;
        if (left == "-") {
            binaryTree[i].left = -1;
        } else {
            int child = atoi(left.c_str());
            children[child] = true;
            binaryTree[i].left = child;
        }
        if (right == "-") {
            binaryTree[i].right = -1;
        } else {
            int child = atoi(right.c_str());
            children[child] = true;
            binaryTree[i].right = child;
        }
    }
    for (int i = 0; i < children.size(); i++) {
        if (!children[i]) {
            root = i;
            break;
        }
    }
    invert(root);
    levelOrderTraverse();
    printf("\n");
    inOrderTraverse(root);
    return 0;
}