二叉樹祖先節(jié)點

Problem statement:

問題陳述：

Given a Binary Tree and a target key, write a function that prints all the ancestors of the key in the given binary tree.

給定二叉樹和目標鍵，編寫一個函數(shù)，以打印給定二叉樹中鍵的所有祖先 。

Example:

例：

Let's the tree be like following:

讓樹如下所示：

Let for node value 12:Ancestors are:7, 5, 8While for node value 7:Ancestors are:5, 8

Solution

解

What is Ancestors?

什么是祖先？

For any node n,
Its ancestors are the nodes which are on the path between roots to node n

對于任何節(jié)點n ，
它的祖先是位于根到節(jié)點n之間的路徑上的節(jié)點

Thus for the above examples,

因此，對于以上示例，

Example 1:

范例1：

Node is 12 //represented by valueRoot to the node path is8->5->7->12Thus the ancestors are 7, 5, 8

Example 2:

范例2：

Node is 7 //represented by valueRoot to the node path is8->5->7Thus the ancestors are 5, 8

Algorithm:

算法：

FUNCTION printAncestors(Node *root, int target)IF(!root)return false;IF( (root->left && root->left->data==target) ||(root->right && root->right->data==target ) || printAncestors(root->left,target)|| printAncestors(root->right,target)){Print root->data;return true;END IFreturn false; END FUNCTION

That simply means we are doing kind of DFS

這僅表示我們正在執(zhí)行某種DFS

For a currentnode to be ancestor of the target node the conditions are:

為了使currentnode成為目標節(jié)點的祖先，條件是：

1. If the target node is its child node (either left child or right child)//conditionIF((root->left && root->left->data==target) ||(root->right && root->right->data==target ))2. If any of the two subtree of the current node contain ancestor of the target node then the current node is also an ancestor. //conditionIF(printAncestors(root->left, target)|| printAncestors(root->right, target))

Example with explanation:

帶有說明的示例：

For target node 7: Root 8 is ancestor on condition: its left subtree contains ancestor 5 5 is ancestor since target node is its right child Thus ancestors are: 5, 8 //in order .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}} .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}}

C++ Implementation:

C ++實現(xiàn)：

#include <bits/stdc++.h> using namespace std;//tree node class Node{ public:int data;Node *left;Node *right; };bool printAncestors(Node *root, int target) {if(!root)return false;if( (root->left && root->left->data==target) ||(root->right && root->right->data==target ) || printAncestors(root->left,target)|| printAncestors(root->right,target)){cout<<root->data<<" ";return true;}return false; }//creating new nodes Node* newnode(int data){ Node* node = (Node*)malloc(sizeof(Node)); node->data = data; node->left = NULL; node->right = NULL; return(node); } int main() { //**same tree is builted as shown in example**cout<<"tree in the example is build here"<<endl;//building the tree like as in the exampleNode *root=newnode(8); root->left= newnode(5); root->right= newnode(4); root->right->right=newnode(11);root->right->right->left=newnode(3);root->left->left=newnode(9); root->left->right=newnode(7);root->left->right->left=newnode(1);root->left->right->right=newnode(12);root->left->right->right->left=newnode(2);int s;cout<<"enter input value to find ancestors......"<<endl;cin>>s;printAncestors(root,s);return 0; }

Output

輸出量

tree in the example is build here enter input value to find ancestors...... 7 5 8

翻譯自: https://www.includehelp.com/icp/ancestors-in-binary-tree.aspx

二叉樹祖先節(jié)點

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