力扣算法题—111.Minimum Depth of Binary Tree

 

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.

Solution:

　　使用深度遍历和广度遍历

class Solution {
private:
    int minLevel = INT32_MAX;
public:
    int minDepth(TreeNode* root) {
        if (root == nullptr)return 0;
        //return BFS(root);
        int res = INT32_MAX;
        DFS(root, 1, res);
        return res;
    }

    int BFS(TreeNode* root)
    {
        queue<TreeNode*>q;
        q.push(root);
        int level = 0;
        while (!q.empty())
        {
            queue<TreeNode*>temp;
            ++level;
            while (!q.empty())
            {
                TreeNode* p = q.front();
                q.pop();
                if (p->left == nullptr && p->right == nullptr)return level;
                if (p->left != nullptr)temp.push(p->left);
                if (p->right != nullptr)temp.push(p->right);
            }
            q = temp;
        }
        return level;
    }
    void DFS(TreeNode* root, int level, int &res)
    {
        if (root->left == nullptr && root->right == nullptr) {
            res = res > level ? level : res;
            return;
        }
        if (root->left != nullptr)DFS(root->left, level + 1, res);
        if (root->right != nullptr)DFS(root->right, level + 1, res);
    }

};