求树的最小深度 

 1 package algo;
 2 
 3 import java.util.LinkedList;
 4 import java.util.Queue;
 5 import java.lang.Math;
 6 
 7 /*
 8  * Description:
 9  * Given a binary tree, find its minimum depth.The minimum depth is the number of
10  *  nodes along the shortest path from the root node down to the nearest leaf node.
11  */
12 class TreeNode {
13     int val;
14     TreeNode left;
15     TreeNode right;
16 
17     TreeNode(int x) {
18         val = x;
19     }
20 }
21 
22 public class Minimum_depth_of_binary_tree {
23     /*
24      * 非递归，广度优先遍历，通过队列遍历每层节点，遇到的第一个叶节点即为最小深度的叶节点。
25      */
26     public static int run(TreeNode root) {
27         int minDepth = 0;
28         Queue<TreeNode> q = new LinkedList<TreeNode>();
29         if (root != null) {
30             q.offer(root);
31         }
32         while (!q.isEmpty()) {
33             int len = q.size();
34             minDepth++;
35             for (int i = 0; i < len; i++) {
36                 TreeNode tmp = q.poll();
37                 if (tmp.left != null) {
38                     q.offer(tmp.left);
39                 }
40                 if (tmp.right != null) {
41                     q.offer(tmp.right);
42                 }
43                 if (tmp.left == null && tmp.right == null) {
44                     return minDepth;
45                 }
46             }
47         }
48         return minDepth;
49     }
50 
51     /*
52      * 递归，深度优先遍历，问题可以归结为：当前节点下最小深度=min(左子树最小深度,右子树最小深度)+1
53      * 递归终止条件为root==null。
54      */
55     public static int run_1(TreeNode root) {
56         if (root == null) {
57             return 0;
58         }
59         int l = run_1(root.left);
60         int r = run_1(root.right);
61         if (l == 0 || r == 0) {
62             return l + r + 1;
63         }
64         return 1 + Math.min(l, r);
65     }
66 
67     public static void main(String[] args) {
68         TreeNode treeNode = new TreeNode(1);
69         treeNode.left = new TreeNode(2);
70         treeNode.left.left = new TreeNode(3);
71         treeNode.left.left.left = new TreeNode(4);
72         System.out.println(Minimum_depth_of_binary_tree.run(treeNode));
73         System.out.println(Minimum_depth_of_binary_tree.run_1(treeNode));
74     }
75 }

 运行结果：

4

4

posted on 2017-08-07 11:03  Bro.Yi  阅读(233)  评论(0)    收藏  举报

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