在 C++ 程式中查詢二叉樹中兩個節點之間的距離
在這個問題中,我們給定一棵二叉樹和兩個節點。我們的任務是建立一個程式來查詢二叉樹中兩個節點之間的距離。
問題描述
我們需要找到兩個節點之間的距離,即從一個節點到另一個節點遍歷的最小邊數。
讓我們舉個例子來理解這個問題,
輸入:二叉樹
節點1 = 3,節點2 = 5
輸出: 3
解釋
從節點 3 到節點 5 的路徑是 3 -> 1 -> 2 -> 5。遍歷了 3 條邊,距離為 3。
解決方案
解決此問題的一個簡單方法是使用給定節點的最近公共祖先節點,然後應用以下公式:
distance(node1, node2) = distance(root, node1) + distance(root, node2) + distance(root, LCA)
示例
#include <iostream> using namespace std; struct Node{ struct Node *left, *right; int key; }; Node* insertNode(int key){ Node *temp = new Node; temp->key = key; temp->left = temp->right = NULL; return temp; } int calcNodeLevel(Node *root, int val, int level) { if (root == NULL) return -1; if (root->key == val) return level; int lvl = calcNodeLevel(root->left, val, level+1); return (lvl != -1)? lvl : calcNodeLevel(root->right, val, level+1); } Node *findDistanceRec(Node* root, int node1, int node2, int &dist1, int &dist2, int &dist, int lvl){ if (root == NULL) return NULL; if (root->key == node1){ dist1 = lvl; return root; } if (root->key == node2){ dist2 = lvl; return root; } Node *leftLCA = findDistanceRec(root->left, node1, node2, dist1,dist2, dist, lvl+1); Node *rightLCA = findDistanceRec(root->right, node1, node2, dist1,dist2, dist, lvl+1); if (leftLCA && rightLCA){ dist = dist1 + dist2 - 2*lvl; return root; } return (leftLCA != NULL)? leftLCA: rightLCA; } int CalcNodeDistance(Node *root, int node1, int node2) { int dist1 = -1, dist2 = -1, dist; Node *lca = findDistanceRec(root, node1, node2, dist1, dist2, dist, 1); if (dist1 != -1 && dist2 != -1) return dist; if (dist1 != -1){ dist = calcNodeLevel(lca, node2, 0); return dist; } if (dist2 != -1){ dist = calcNodeLevel(lca, node1, 0); return dist; } return -1; } int main(){ Node * root = insertNode(1); root->left = insertNode(2); root->right = insertNode(3); root->left->left = insertNode(4); root->left->right = insertNode(5); root->right->left = insertNode(6); cout<<"Distance between node with value 5 and node with value 3 is"<<CalcNodeDistance(root, 3, 5); return 0; }
輸出
Distance between node with value 5 and node with value 3 is 3
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