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111. Minimum Depth of Binary Tree

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.

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

一刷，按层判断

 1 class Solution(object):
 2     def minDepth(self, root):
 3         if not root:
 4             return 0
 5         
 6         prev = [root]
 7         cur = []
 8         depth = 1
 9         while prev:
10             for elem in prev:
11                 if not elem.left and not elem.right:
12                     return depth
13                 if elem.left:
14                     cur.append(elem.left)
15                 if elem.right:
16                     cur.append(elem.right)
17             depth += 1
18             prev, cur = cur, []
19         return depth

2/17/2017, Java, performance不好

错误：判断时候要注意是否是叶节点，而不是一边是null，一边有子树

 1 public class Solution {
 2     public int minDepth(TreeNode root) {
 3         if (root == null) return 0;
 4         Queue<TreeNode> level = new LinkedList<>();
 5         TreeNode node = new TreeNode(-1);
 6         int size;
 7         int index = 0;
 8         
 9         level.add(root);
10         while(!level.isEmpty()) {
11             index++;
12             size = level.size();
13             for (int i = 0; i < size; i++) {
14                 node = level.poll();
15                 if (node.left == null && node.right == null) return index;
16                 if (node.left != null) level.add(node.left);
17                 if (node.right != null) level.add(node.right);
18             }
19         }
20         return index;
21     }
22 }

5/11/2017

算法班

注意：

1. Queue是接口，用LinkedList实现

2. Queue的方法是add/poll/remove等等

3. 判断LinkedList是否为空，isEmpty与stack.empty有不同

 1 public class Solution {
 2     public int minDepth(TreeNode root) {
 3         if (root == null) return 0;
 4         
 5         Queue<TreeNode> queue = new LinkedList<TreeNode>();
 6         TreeNode node;
 7         queue.add(root);
 8         int index = 0;
 9         while (!queue.isEmpty()) {
10             index++;
11             int size = queue.size();
12             for (int i = 0; i < size; i++) {
13                 node = queue.poll();
14                 if (node.left == null && node.right == null) return index;
15                 if (node.left != null) {
16                     queue.add(node.left);
17                 }
18                 if (node.right != null) {
19                     queue.add(node.right);
20                 }
21             }
22         }
23         return index;
24     }
25 }

别人的做法

递归

1 public class Solution {
2     public int minDepth(TreeNode root) {
3         if(root == null) return 0;
4         int left = minDepth(root.left);
5         int right = minDepth(root.right);
6         return (left == 0 || right == 0) ? left + right + 1: Math.min(left,right) + 1;
7        
8     }
9 }

https://discuss.leetcode.com/topic/8723/my-4-line-java-solution

https://discuss.leetcode.com/topic/16869/3-lines-in-every-language

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原文地址：https://www.cnblogs.com/panini/p/5602954.html