LeetCode - Minimum Depth of Binary Tree

原题链接：http://oj.leetcode.com/problems/minimum-depth-of-binary-tree/ 

这道题是树的题目，其实跟Maximum Depth of Binary Tree非常类似，只是这道题因为是判断最小深度，所以必须增加一个叶子的判断（因为如果一个节点如果只有左子树或者右子树，我们不能取它左右子树中小的作为深度，因为那样会是0，我们只有在叶子节点才能判断深度，而在求最大深度的时候，因为一定会取大的那个，所以不会有这个问题）。这道题同样是递归和非递归的解法，递归解法比较常规的思路，比Maximum Depth of Binary Tree多加一个左右子树的判断，代码如下：

//题目描述////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./** * @author Administrator * */public class minimum_depth_of_binarytree {class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int x) {val = x;}}// 递归解法public int myRun(TreeNode root) {if (root == null)return 0;if (root.left == null && root.right == null)return 1;// 叶子节点判断if (root.left == null)return myRun(root.right) + 1;if (root.right == null)return myRun(root.left) + 1;int left = myRun(root.left) + 1;int right = myRun(root.right) + 1;return Math.min(left, right);}// 非递归解法// 非递归解法同样采用层序遍历(相当于图的BFS），// 只是在检测到第一个叶子的时候就可以返回了public int minDepth(TreeNode root) {if (root == null)return 0;LinkedList<TreeNode> queue = new LinkedList<TreeNode>();int curNum = 0;int lastNum = 1;int level = 1;queue.offer(root);while (!queue.isEmpty()) {TreeNode cur = queue.poll();if (cur.left == null && cur.right == null)return level;lastNum--;if (cur.left != null) {queue.offer(cur.left);curNum++;}if (cur.right != null) {queue.offer(cur.right);curNum++;}if (lastNum == 0) {lastNum = curNum;curNum = 0;level++;}}return 0;}/** * @param args */public static void main(String[] args) {// TODO Auto-generated method stub}}