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

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.

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<TreeNode*> 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;

									    }

									};