Find Bottom Left Tree Value: Accepted

Author: gouthampradhan

README.md

@@ -201,6 +201,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Binary Tree Inorder Traversal](problems/src/tree/BinaryTreeInorderTraversal.java) (Medium)
 - [Symmetric Tree](problems/src/tree/SymmetricTree.java) (Easy)
 - [Maximum Binary Tree](problems/src/tree/MaximumBinaryTree.java) (Medium)
+- [Find Bottom Left Tree Value](problems/src/tree/FindBottomLeftTreeValue.java) (Medium)
 
 #### [Two Pointers](problems/src/two_pointers)
 

problems/src/tree/FindBottomLeftTreeValue.java

@@ -0,0 +1,68 @@
+package tree;
+
+/**
+ * Created by gouthamvidyapradhan on 30/08/2017.
+ * Given a binary tree, find the leftmost value in the last row of the tree.
+
+ Example 1:
+ Input:
+
+ 2
+ / \
+ 1   3
+
+ Output:
+ 1
+ Example 2:
+ Input:
+
+ 1
+ / \
+ 2   3
+ /   / \
+ 4   5   6
+ /
+ 7
+
+ Output:
+ 7
+ Note: You may assume the tree (i.e., the given root node) is not NULL.
+
+
+ */
+public class FindBottomLeftTreeValue {
+    private int max = 0, result;
+    public static class TreeNode {
+        int val;
+        TreeNode left;
+        TreeNode right;
+        TreeNode(int x) { val = x; }
+    }
+
+    public static void main(String[] args) throws Exception{
+        TreeNode root = new TreeNode(1);
+        root.left = new TreeNode(2);
+        root.left.left = new TreeNode(4);
+        root.right = new TreeNode(3);
+        root.right.left = new TreeNode(5);
+        root.right.left.left = new TreeNode(7);
+        root.right.right = new TreeNode(6);
+        System.out.println(new FindBottomLeftTreeValue().findBottomLeftValue(root));
+    }
+
+    public int findBottomLeftValue(TreeNode root) {
+        preorder(root, 1);
+        return result;
+    }
+
+    private void preorder(TreeNode node, int level){
+        if(node != null){
+            if(level > max){
+                result = node.val;
+                max = level;
+            }
+            preorder(node.left, level + 1);
+            preorder(node.right, level + 1);
+        }
+    }
+}