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.
Tree Depth-first Search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
更多讨论
https://discuss.leetcode.com/category/119/minimum-depth-of-binary-tree