题目：求二叉树叶子节点的个数

思路：递归和迭代

递归：求叶子节点的个数，就是求 左子树叶子节点的个数  加上  右子树叶子节点的个数

public static int numberOfLeaf(TreeNode root){        if(root == null)            return 0;        if(root.left == null && root.right == null)            return 1;        return numberOfLeaf(root.left)+numberOfLeaf(root.right);}

迭代：

public static int numberOfLeaves(TreeNode root){        if(root == null)            return 0;        if(root.left == null && root.right == null)            return 1;        Queue
     
       queue = new LinkedList
      
       ();        queue.add(root);        int leafNodes = 0;        while(!queue.isEmpty()){            TreeNode cur = queue.remove();            if(cur.left != null){                queue.add(cur.left);            }            if(cur.right != null){                queue.add(cur.right);            }            if(cur.left == null && cur.right == null){                leafNodes++;            }        }        return leafNodes;    }