220429二叉树的最小深度-广度优先 - 图1

    1. public static void main(String[] args) {
    2.         TreeNode treeNode7 = new TreeNode(7,null, null);
    3.         TreeNode treeNode6 = new TreeNode(6, treeNode7, null);
    4.         TreeNode treeNode5 = new TreeNode(5,null, treeNode6);
    5.         TreeNode treeNode4 = new TreeNode(4,null, null);
    6.         TreeNode treeNode3 = new TreeNode(3,null, null);
    7.         TreeNode treeNode2 = new TreeNode(2, treeNode3, treeNode4);
    8.         TreeNode treeNode1 = new TreeNode(1, treeNode2, treeNode5);
    9.         System.out.println(scope(treeNode1));
    11.      }
    13.     private static int scope(TreeNode treeNode) {
    14.         if (null == treeNode) return 0;
    15.         // 定义一个先进先出的队列
    16.         Queue<TreeNode> queue = new LinkedBlockingDeque<TreeNode>();
    18.         // 根节点的深度设置为1
    19.         treeNode.deep = 1;
    20.         queue.offer(treeNode);
    21.         while (null != queue) {
    22.             treeNode = queue.poll();
    23.             if (null == treeNode.left && null == treeNode.right) {
    24.                 // 返回最先遍历到的叶节点的深度
    25.                 return treeNode.deep;
    26.             }
    27.             // 节点的左子节点不为空
    28.             if (null != treeNode.left) {
    29.                 // 左子节点的深度+1
    30.                 treeNode.left.deep = treeNode.deep + 1;
    31.                 // 将左子节点放入队列中
    32.                 queue.offer(treeNode.left);
    33.             }
    34.             // 节点的右子节点不为空
    35.             if (null != treeNode.right) {
    36.                 // 右子节点的深度+1
    37.                 treeNode.right.deep = treeNode.deep + 1;
    38.                 // 将右子节点放入队列中
    39.                 queue.offer(treeNode.right);
    40.             }
    41.         }
    43.         return 0;
    44.     }