C++实现LeetCode(111.二叉树的最小深度)

• [LeetCode] 111. Minimum Depth of Binary Tree 二叉树的最小深度

[LeetCode] 111. Minimum Depth of Binary Tree 二叉树的最小深度

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.

Note: A leaf is a node with no children

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Example:

Given binary tree [3,9,20,null,null,15,7],

    3
/ \
9  20
/  \
15   7

return its minimum depth = 2.

二叉树的经典问题之最小深度问题就是就最短路径的节点个数，还是用深度优先搜索 DFS 来完成，万能的递归啊。首先判空，若当前结点不存在，直接返回0。然后看若左子结点不存在，那么对右子结点调用递归函数，并加1返回。反之，若右子结点不存在，那么对左子结点调用递归函数，并加1返回。若左右子结点都存在，则分别对左右子结点调用递归函数，将二者中的较小值加1返回即可，参见代码如下：

解法一：

 class Solution { public: int minDepth(TreeNode* root) { if (!root) return 0; if (!root->left) return 1 + minDepth(root->right); if (!root->right) return 1 + minDepth(root->left); return 1 + min(minDepth(root->left), minDepth(root->right)); } };

我们也可以是迭代来做，层序遍历，记录遍历的层数，一旦遍历到第一个叶结点，就将当前层数返回，即为二叉树的最小深度，参见代码如下：

解法二：

 class Solution { public: int minDepth(TreeNode* root) { if (!root) return 0; int res = 0; queue q{{root}}; while (!q.empty()) { ++res; for (int i = q.size(); i > 0; --i) { auto t = q.front(); q.pop(); if (!t->left && !t->right) return res; if (t->left) q.push(t->left); if (t->right) q.push(t->right); } } return -1; } };

Github 同步地址：

https://github.com/grandyang/leetcode/issues/111

类似题目：

Binary Tree Level Order Traversal

Maximum Depth of Binary Tree

参考资料：

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

https://leetcode.com/problems/minimum-depth-of-binary-tree/discuss/36153/My-concise-c%2B%2B-solution

https://leetcode.com/problems/minimum-depth-of-binary-tree/discuss/36071/BFS-C%2B%2B-8ms-Beats-99.94-submissions