Determine whether a tree is a complete binary tree
2023 01 10 107 00 46 07
I won’t talk about the definition of a complete binary tree, so how do we judge whether a tree is an ordinary tree or a complete binary tree?
Title description :
Given the head node TreeNode root of a binary tree, it is required to return true or false.
Code :
In fact, it is the idea of using layer order traversal
boolean isCompleteTree(TreeNode root){
if(root == null){
return true;
}
Deque<TreeNode> deque = new LinkedList<>();
deque.offer(root);
while(!deque.isEmpty()){
TreeNode cur = deque.poll();
if(cur != null){
deque.offer(cur.left);
deque.offer(cur.right);
}
else{
break;
}
}
while(!deque.isEmpty()){
TreeNode cur = deque.poll();
if(cur != null){
return false;
}
}return true;
}
Definition of the test tree
ublic class MyBinaryTree {
class TreeNode {
char val;
TreeNode left;
TreeNode right;
public TreeNode(char val) {
this.val = val;
this.left = null;
this.right = null;
}
}
public TreeNode root;//根节点
}
Test class
public class Test {
public static void main(String[] args) {
MyBinaryTree myBinaryTree1 = new MyBinaryTree();
MyBinaryTree.TreeNode root = myBinaryTree1.root;
MyBinaryTree.TreeNode A = new MyBinaryTree().new TreeNode('A');
MyBinaryTree.TreeNode B = new MyBinaryTree().new TreeNode('B');
MyBinaryTree.TreeNode C = new MyBinaryTree().new TreeNode('C');
MyBinaryTree.TreeNode D = new MyBinaryTree().new TreeNode('D');
MyBinaryTree.TreeNode E = new MyBinaryTree().new TreeNode('E');
MyBinaryTree.TreeNode F = new MyBinaryTree().new TreeNode('F');
MyBinaryTree.TreeNode G = new MyBinaryTree().new TreeNode('G');
MyBinaryTree.TreeNode H = new MyBinaryTree().new TreeNode('H');
A.left = B;A.right = C;B.left = D;B.right = E;E.right = H;C.left = F;C.right = G;
root = A;
//判断是不是完全二叉树
System.out.println("判断是不是完全二叉树");
System.out.println(myBinaryTree1.isCompleteTree(root));
}
}
result