【剑指offer-Java版】23从上往下打印二叉树

谁践踏了优雅 2022-08-21 09:40 98阅读 0赞

从上往下打印二叉树:如果把这个书看做是一个特殊的图，那么该打印过程就类似于一个广度优先遍历

public class _Q23 {
    
        public <T> void PrintTreeFromTopToButtom(BinaryTreeNode<T> root){
            if(root == null) return;
    
            Queue<BinaryTreeNode<T>> queue = new LinkedList<>();
    
            queue.add(root);
            while(!queue.isEmpty()){
                BinaryTreeNode<T> node = queue.remove();
    
                if(node.leftChild != null) queue.add(node.leftChild);
                if(node.rightChild != null) queue.add(node.rightChild);
    
                System.out.print(node.value + " ");
            }
            System.out.println();
        }
        }

测试代码：

public class _Q23Test extends TestCase {
        
    
        _Q23 tree = new _Q23();
    
        public void test(){
            BinaryTreeNode<Integer> root = new BinaryTreeNode<>();
            BinaryTreeNode<Integer> node1 = new BinaryTreeNode<>();
            BinaryTreeNode<Integer> node2 = new BinaryTreeNode<>();
            BinaryTreeNode<Integer> node3 = new BinaryTreeNode<>();
            BinaryTreeNode<Integer> node4 = new BinaryTreeNode<>();
            BinaryTreeNode<Integer> node5 = new BinaryTreeNode<>();
            BinaryTreeNode<Integer> node6 = new BinaryTreeNode<>();
    
            root.value = 8;
            node1.value = 6;
            node2.value = 10;
            node3.value = 5;
            node4.value = 7;
            node5.value = 9;
            node6.value = 11;
    
            root.leftChild = node1;
            root.rightChild = node2;
            node1.leftChild = node3;
            node1.rightChild = node4;
            node2.leftChild = node5;
            node2.rightChild = node6;
    
            node3.leftChild = null; node3.rightChild = null;
            node4.leftChild = null; node4.rightChild = null;
            node5.leftChild = null; node5.rightChild = null;
            node6.leftChild = null; node6.rightChild = null;
    
    
            tree.PrintTreeFromTopToButtom(root);
            tree.PrintTreeFromTopToButtom(null);
            tree.PrintTreeFromTopToButtom(node1);
        }
        }