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题目要求:
如标题:
分析:
对于二叉树中的任意一个节点,层序遍历二叉树①:如果只有右孩子,没有左孩子,那么该树一定不是完全二叉树;②:第一次发现一个节点有如果左没右,或者左右都没有,那么其后的每一个节点都必须是叶子节点,否则一定不是一颗完全二叉树;
package com.isea.brush.tree;
import java.util.LinkedList;
import java.util.Queue;
/**
* 判断一棵树是否是一颗完全二叉树
*/
public class IsCBT {
private static class Node {
private Integer data;
private Node left;
private Node right;
public Node(int data) {
this.data = data;
}
}
public static boolean isCBT(Node head) {
if (head == null) {
return true;
}
Queue<Node> queue = new LinkedList<Node>();
boolean leaf = false;
Node left = null;
Node right = null;
queue.offer(head);
while (!queue.isEmpty()) {
head = queue.poll();
left = head.left;
right = head.right;
if ((leaf && (left != null || right != null)) || (left == null && right != null)) {
return false;
}
if (left != null){
queue.offer(left);
}
if (right != null){
queue.offer(right);
}
if (left == null || right == null){
leaf = true;
}
}
return true;
}
public static void main(String[] args) {
Node head = new Node(4);
head.left = new Node(2);
head.right = new Node(6);
head.left.left = new Node(1);
// head.left.right = new Node(3);
head.right.left = new Node(5);
System.out.println(isCBT(head)); // true
}
}