C++ level order traversal solution
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n/**\n * Definition for a binary tree node.\n * struct TreeNode {\n *     int val;\n *     TreeNode *left;\n *     TreeNode *right;\n *     TreeNode() : val(0), left(nullptr), right(nullptr) {}\n *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}\n *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}\n * };\n */\nclass Solution {\npublic:\n    int findBottomLeftValue(TreeNode* root) {\n        queue<pair<TreeNode*,int>>q;\n        q.push({root, 0});\n        int ans = 0;\n        int hl = INT_MAX;\n\n        while(!q.empty()){\n            int size = q.size();\n            ans = q.front().first->val;\n            for(int i=0;i<size;i++){\n                TreeNode* currNode = q.front().first;\n                int currHl = q.front().second;\n                q.pop();\n                if(currHl<hl){\n                    hl = currHl;\n                    ans = currNode->val;\n                }\n                TreeNode* left = currNode->left;\n                TreeNode* right = currNode->right;\n                if(left != NULL){\n                    q.push({left, currHl-1});\n                }\n                if(right != NULL){\n                    q.push({right, currHl+1});\n                }\n            }\n            \n        }\n        return ans;\n    }\n};\n```
['C++']