Easy solution using java
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:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->O(N) for N nodes\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->O(W) for W=> widht of tree(max)\n\n# Code\n```java []\n/**\n * Definition for a binary tree node.\n * public class TreeNode {\n *     int val;\n *     TreeNode left;\n *     TreeNode right;\n *     TreeNode() {}\n *     TreeNode(int val) { this.val = val; }\n *     TreeNode(int val, TreeNode left, TreeNode right) {\n *         this.val = val;\n *         this.left = left;\n *         this.right = right;\n *     }\n * }\n */\nclass Solution {\n    public int findBottomLeftValue(TreeNode root) {\n        Queue<TreeNode> q= new LinkedList<>();\n\n        q.add(root);\n        \n\n        int leftBottomValue=root.val;\n\n        while(!q.isEmpty()){\n\n            int levelSize=q.size();\n\n            for(int i=0;i<levelSize;i++){\n                TreeNode curr=q.remove();\n\n\n                if(i==0){ \n                    leftBottomValue=curr.val;  // it will add very first Node at every level to it,and end up adding the leftMost value of bottom level\n\n                }\n\n                if(curr.left!=null){\n                    q.add(curr.left);\n                }\n                if(curr.right!=null){\n                    q.add(curr.right);\n                }\n            }\n        \n        }\n        return leftBottomValue;\n    }\n}\n\n\n/**\n\nif ques is for bottom right then simple check conditions with\nif(i==size-1) */\n```
['Java']