Create 0662-maximum-width-of-binary-tree.kt

Author: a93a

Diff for: kotlin/0662-maximum-width-of-binary-tree.kt

@@ -0,0 +1,69 @@
+/**
+ * Example:
+ * var ti = TreeNode(5)
+ * var v = ti.`val`
+ * Definition for a binary tree node.
+ * class TreeNode(var `val`: Int) {
+ *     var left: TreeNode? = null
+ *     var right: TreeNode? = null
+ * }
+ */
+
+/*
+* BFS solution
+*/
+class Solution {
+    fun widthOfBinaryTree(root: TreeNode?): Int {
+        root?: return 0
+
+        val q = ArrayDeque<Triple<TreeNode?, Int, Int>>()
+        
+        var res = 0
+        var prevLevel = 0
+        var prevNum = 1
+        q.add(Triple(root, 1, 0))
+
+        while (q.isNotEmpty()) {
+            val (node, num, level) = q.poll()
+
+            if (level > prevLevel) {
+                prevLevel = level
+                prevNum = num
+            }
+
+            res = maxOf(res, num - prevNum + 1)
+
+            node?.left?.let {
+                q.add(Triple(node.left, 2 * num, level + 1)) 
+            }
+            node?.right?.let {
+                q.add(Triple(node.right, 2 * num + 1, level + 1)) 
+            }
+        }
+
+        return res
+    }   
+}
+
+/*
+* DFS solution
+*/
+class Solution {
+    fun widthOfBinaryTree(root: TreeNode?): Int {
+        var width = 0
+        val levelMap = HashMap<Int, Int>()
+        
+        fun dfs(node: TreeNode?, depth: Int, pos: Int, levelMap: HashMap<Int, Int>) {
+            node?: return
+            if(!levelMap.contains(depth))
+                levelMap[depth] = pos
+            width = maxOf(width, (pos - levelMap[depth]!! + 1))
+            dfs(node.left, depth+1, 2*pos, levelMap)
+            dfs(node.right, depth+1, 2*pos+1, levelMap)
+        }
+              
+        dfs(root, 0, 0, levelMap)
+        return width
+    }
+    
+}