Added a new C++ Program

DSA Problems 7 Binary Tree/Validate_binary_tree_nodes.cpp

@@ -0,0 +1,48 @@
+/*
+1361. Validate Binary Tree Nodes
+
+Problem Link:- https://leetcode.com/problems/validate-binary-tree-nodes/?envType=daily-question&envId=2023-10-17
+
+You have n binary tree nodes numbered from 0 to n - 1 where node i has two children leftChild[i] and rightChild[i], return true if and only if all the given nodes form exactly one valid binary tree.
+
+If node i has no left child then leftChild[i] will equal -1, similarly for the right child.
+
+Note that the nodes have no values and that we only use the node numbers in this problem.
+
+Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,3,-1,-1]
+Output: false
+
+*/
+
+// SOLUTION FUNCTION
+class Solution {
+public:
+    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
+        vector<int>indegree(n,0);
+        for(int i=0;i<n;i++){
+            if(leftChild[i]!=-1) indegree[leftChild[i]]++;
+            if(rightChild[i]!=-1) indegree[rightChild[i]]++;
+        }
+        int root=-1;
+        for(int i=0;i<n;i++){
+            if(indegree[i]==0){
+                if(root==-1) root=i;
+                else
+                    return false;
+            }
+        }
+        if(root==-1) return false;
+        vector<bool>visited(n,false);
+        queue<int>q;
+        q.push(root);
+        while(!q.empty()){
+            int node=q.front();
+            q.pop();
+            if(visited[node]) return false;
+            visited[node]=true;
+            if(leftChild[node]!=-1) q.push(leftChild[node]);
+            if(rightChild[node]!=-1) q.push(rightChild[node]);
+        }
+        return accumulate(visited.begin(),visited.end(),0)==n;
+    }
+};