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 ;
         //return BFS(root);
         int res = INT32_MAX;
         DFS(root, , res);
         return res;
     }

     int BFS(TreeNode* root)
     {
         queue<TreeNode*>q;
         q.push(root);
         int level = ;
         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 + , res);
         if (root->right != nullptr)DFS(root->right, level + , res);
     }

 };