111. Minimum Depth of Binary Tree (最小树深度\bfs)

 

 

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.

 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:

    int minDepth(TreeNode* root) {
        if (root==nullptr) return 0;
        queue<TreeNode*> q;
        q.push(root);
        int depth = 1;
        while(!q.empty()) {
            int sz = q.size();
            for(int i = 0; i < sz;i++) {
                TreeNode* cur = q.front(); q.pop();
                if (cur->left == nullptr && cur->right == nullptr) {
                    return depth;
                }
                if (cur->left != nullptr) q.push(cur->left);
                if (cur->right!= nullptr) q.push(cur->right);
            }
            depth++;
        }
        return depth;
    }
};

 

 

 

 1 class Solution {
 2 public:
 3 
 4     int minDepth(TreeNode* root) {
 5         if (root == nullptr) return 0;
 6         if (root->left != nullptr && root->right != nullptr) {
 7             return 1 + min(minDepth(root->left),minDepth(root->right));
 8         } else if (root->left != nullptr && root->right == nullptr) {
 9             return 1 + minDepth(root->left);
10         } else if (root->left == nullptr && root->right != nullptr) {
11             return 1 + minDepth(root->right);
12         } 
13         return 1;
14     }
15 };