找到 C++ 中二叉树中给定节点的镜像
在这个问题中,我们给定一棵二叉树。我们的任务是找到二叉树中给定节点的镜像。我们将获得一个节点,并在相反的子树中找到该节点的镜像。
我们举个例子来理解这个问题,
输入
输出
mirror of B is E.
解决方案方法
解决此问题的一个简单解决方案是使用来自根部的递归,并使用两个指针表示左子树和右子树。然后对于目标值,如果找到镜像,则返回镜像,否则递归其他节点。
程序说明我们解决方案的工作原理,
示例
#include <bits/stdc++.h> using namespace std; struct Node { int key; struct Node* left, *right; }; struct Node* newNode(int key){ struct Node* n = (struct Node*) malloc(sizeof(struct Node*)); if (n != NULL){ n->key = key; n->left = NULL; n->right = NULL; return n; } else{ cout << "Memory allocation failed!" << endl; exit(1); } } int mirrorNodeRecur(int node, struct Node* left, struct Node* right){ if (left == NULL || right == NULL) return 0; if (left->key == node) return right->key; if (right->key == node) return left->key; int mirrorNode = mirrorNodeRecur(node, left->left, right->right); if (mirrorNode) return mirrorNode; mirrorNodeRecur(node, left->right, right->left); } int findMirrorNodeBT(struct Node* root, int node) { if (root == NULL) return 0; if (root->key == node) return node; return mirrorNodeRecur(node, root->left, root->right); } int main() { struct Node* root = newNode(1); root-> left = newNode(2); root->left->left = newNode(3); root->left->left->left = newNode(4); root->left->left->right = newNode(5); root->right = newNode(6); root->right->left = newNode(7); root->right->right = newNode(8); int node = root->left->key; int mirrorNode = findMirrorNodeBT(root, node); cout<<"The node is root->left, value : "<<node<<endl; if (mirrorNode) cout<<"The Mirror of Node "<<node<<" in the binary tree is Node "<<mirrorNode; else cout<<"The Mirror of Node "<<node<<" in the binary tree is not present!"; node = root->left->left->right->key; mirrorNode = findMirrorNodeBT(root, node); cout<<"\n\nThe node is root->left->left->right, value : "<<node<<endl; if (mirrorNode) cout<<"The Mirror of Node "<<node<<" in the binary tree is Node "<<mirrorNode; else cout<<"The Mirror of Node "<<node<<" in the binary tree is not present!"; }
输出
The node is root->left, value : 2 The Mirror of Node 2 in the binary tree is Node 6 The node is root->left->left->right, value : 5 The Mirror of Node 5 in the binary tree is not present!
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