标签：for hash hashset multiple const struct set eve only
https://www.careercup.com/question?id=5103530547347456
Given a list of nodes, each with a left child and a right child (they can be null), determine if all nodes belong in a single valid binary tree. The root is not given.
Li家
The solution can be designed with the following idea which runs in multiple passes:
Pass1: Read every node and for each node remember what other nodes are pointing to it using additional data structures
Pass2: Read every node to find the following:
a. Nodes who are pointed to by 0 other nodes ==> These are potential roots
b. Nodes who are pointed to by 1 nodes
c. Nodes who are pointed to by > 1 nodes
We have a valid binary tree iff:
1. The number of nodes whom nobody points to is 1 and that is the root
2. Every node is pointed to by at most one node
3. Starting with the root, and doing a DFS or a BFS covers all the nodes in the list
Java:
public boolean isValid(List<TreeNode> nodes){
HashSet<TreeNode> children = new HashSet<> ();
// child node only has one parent node
for (TreeNode node : nodes) {
if (node.left != null) {
if (!children.add(node.left)) return false ;
}
if (node.right != null) {
if (!children.add(node.right)) return false ;
}
}
TreeNode start = null ;
int count = 0 ;
for (TreeNode node : nodes) {
if (!children.contains(node)) {
start = node ;
count ++ ;
}
}
// only has one root node
if (count > 1) return false ;
// running bfs to make sure all nodes can be constructed as a binary tree
Queue<TreeNode> q = new LinkedList<> ();
q.add(start) ;
while (!q.isEmpty()) {
int size = q.size() ;
for (int i = 0 ; i < size ; ++i) {
TreeNode cur = q.poll() ;
if (cur.left != null) {
q.add(cur.left) ;
children.remove(cur.left) ;
}
if (cur.right != null) {
q.add(cur.right) ;
children.remove(cur.right) ;
}
}
}
return children.size() == 0 ;
}
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[LeetCode] 261. Graph Valid Tree 图是否是树
标签：for hash hashset multiple const struct set eve only
原文地址：https://www.cnblogs.com/lightwindy/p/9744071.html
