leetcode 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.

递归：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        if(root == NULL)
            return 0;
        if(root->left == NULL && root->right == NULL)
            return 1;
        if(root->left == NULL && root->right != NULL)
            return minDepth(root->right) + 1;
        if(root->left != NULL && root->right == NULL)
            return minDepth(root->left) + 1;
        else
            return min(minDepth(root->left), minDepth(root->right)) + 1;
        // if(root == NULL)
        //     return 0;
        // int l = minDepth(root->left);
        // int r = minDepth(root->right);
        // if(l == 0)
        //     return r + 1;
        // if(r == 0)
        //     return l + 1;
        // return (l < r ? l : r) + 1;
    }
};

bfs：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        if(root == NULL)
            return 0;
        int num = 0;
        queue<TreeNode*> q;
        TreeNode* temp;
        q.push(root);
        while(!q.empty()){
            num++;
            int len = q.size();
            while(len--){
                temp = q.front();
                q.pop();
                if(temp->left == NULL && temp->right == NULL)
                    return num;
                if(temp->left != NULL)
                    q.push(temp->left);
                if(temp->right != NULL)
                    q.push(temp->right);
            }
        }
    }
};