原题目：Given a binary tree, find its minimum depth.The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
给定一个二叉树，求最小深度。最小深度即为根节点至最近叶子节点经过的节点数。经过一番挣扎，最终用两个队列交叉使用来解决，实际上就是一个层次遍历，本机测试还可以，在线编程环境不知道为什么会出错。下边代码中的测试用例为：

测试所用二叉树.png

    import java.util.Queue;
    import java.util.LinkedList;
    
    public class HelloWorld {
        public static void main(String[] args){
            TreeNode root = new TreeNode(1);
            TreeNode node2 = new TreeNode(2);
            TreeNode node3 = new TreeNode(3);
            TreeNode node4 = new TreeNode(4);
            TreeNode node5 = new TreeNode(5);
            TreeNode node6 = new TreeNode(6);
            TreeNode node7 = new TreeNode(7);
            TreeNode node8 = new TreeNode(8);
            TreeNode node9 = new TreeNode(9);
    
            root.left = node2;
            root.right = node3;
    
            node2.left=node4;
            node2.right=node5;
    
            node3.left=node6;
            node3.right=node7;
    
            node4.right=node8;
    
            node6.left=node9;
    
            Solution s = new Solution();
            System.out.println(s.run(root));
        }
        static class TreeNode {
            int val;
            TreeNode left;
            TreeNode right;
            TreeNode(int x) { val = x; }
        }
    
        static class Solution {
            private int minimum = 1;
            private Queue<TreeNode> queue1 = new LinkedList<TreeNode>();
            private Queue<TreeNode> queue2 = new LinkedList<TreeNode>();
    
            public int run(TreeNode root) {
                if (root == null) return 0;
                queue1.offer(root);
                while(queue1.size()>0 || queue2.size()>0){
                    if(queue1.size()>0){ 
                        while(queue1.size()>0){//此处不能用null来判断，各位一想便知了
                            TreeNode t = queue1.poll();
                            if(t.left==null && t.right==null) return minimum;
                            if(t.left!=null) queue2.offer(t.left);
                            if(t.right!=null) queue2.offer(t.right);
                        }
                        minimum++;
                    }
                    if(queue2.size()>0){
                        while(queue2.size()>0){//此处不能用null来判断，各位一想便知了
                            TreeNode t = queue2.poll();
                            if(t.left==null && t.right==null) return minimum;
                            if(t.left!=null) queue1.offer(t.left);
                            if(t.right!=null) queue1.offer(t.right);
                        }
                        minimum++;
                    }
                }
                return -1;
            }
        }
    }

代码输出结果:3

发现有问题可随时指出，谢谢。（第一篇博客，做个爱记笔记的好孩子）