【LeetCode 111_二叉树_遍历】Minimum Depth of Binary Tree

解法一:递归

int minDepth(TreeNode* root) 
{
    if (root == NULL)
        return 0;

    if (root->left == NULL) {
        return minDepth(root->right) + 1;
    } else if (root->right == NULL) {
        return minDepth(root->left) + 1;
    } else {
        int left_minDepth = minDepth(root->left);
        int right_minDepth = minDepth(root->right);
        return left_minDepth < right_minDepth ? left_minDepth + 1 : right_minDepth + 1;
    }
}

解法二:BFS

int minDepth(TreeNode *root) 
{
    if (root == NULL)
        return 0;

    queue<TreeNode*> node_queue;
    node_queue.push(root);

    int count = 0;
    while (!node_queue.empty()) {
        int len = node_queue.size();
        count++;
        for (int i = 0; i < len; ++i) {
            TreeNode *nodeTmp = node_queue.front();
            node_queue.pop();

            if (nodeTmp->left)
                node_queue.push(nodeTmp->left);
            if (nodeTmp->right)
                node_queue.push(nodeTmp->right);
            if (nodeTmp->left == NULL && nodeTmp->right == NULL)
                return count;
        }
    }
}
原文地址:https://www.cnblogs.com/mengwang024/p/4621374.html