二叉树简单题：求二叉树节点个数/求二叉树叶子节点的个数/求二叉树第k层节点的个数/在二叉树中查找指定的元素

梦里梦外; 2023-01-10 14:40 106阅读 0赞

package day_21_1_24;
    
    class Node {
        
        public char val;
        public Node left;
        public Node right;
    
        public Node(char val) {
        
            this.val = val;
        }
    }
    public class TestTree {
        
        public static Node build(){
        
            //手动把一颗树构造出来
            Node a = new Node('A');
            Node b = new Node('B');
            Node c = new Node('C');
            Node d = new Node('D');
            Node e = new Node('E');
            Node f = new Node('F');
            Node g = new Node('G');
            Node h = new Node('H');
    
            a.left = b;
            a.right = c;
            b.left = d;
            b.right = e;
            e.left = g;
            g.right = h;
            c.right = f;
    
            //返回根节点
            return a;
        }
    
        //求二叉树节点个数
        //使用后序遍历
        public static int size(Node root){
        
            if (root == null){
        
                return 0;
            }
            //访问根节点，此处的访问操作就是计数器+1
            //整个树的节点个数 = 根节点个数 + 左子树节点的个数 + 右子树节点个数
            return 1 + size(root.left) + size(root.right);
        }
    
        //求二叉树叶子节点的个数
        //root叶子节点的个数 = root.left的叶子节点的个数 + root.right叶子节点的个数
        public static int leafSize(Node root){
        
            if (root == null){
        
                return 0;
            }
            if (root.left == null && root.right == null){
        
                return 1;
            }
            return leafSize(root.left) + leafSize(root.right);
        }
    
        //求二叉树第k层节点的个数
        public static int kSize(Node root,int k){
        
            if (k < 1){
        
                return 0;
            }
            if (k == 1){
        
                return 1;
            }
            return kSize(root.left, k-1)+kSize(root.right, k-1);
        }
    
        //在二叉树中查找指定的元素
        public static Node search(Node root, char toFind){
        
            if (root == null){
        
                return null;
            }
            if (root.val == toFind){
        
                return root;
            }
            Node result = search(root.left, toFind);
            if (result != null){
        
                return result;
            }
            return search(root.right, toFind);
        }
    
    }