Reorder List/ Binary Tree Preorder Traversal

Author: gavinfish

143 Reorder List.py

@@ -0,0 +1,50 @@
+'''
+Given a singly linked list L: L0→L1→…→Ln-1→Ln,
+reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
+
+You must do this in-place without altering the nodes' values.
+
+For example,
+Given {1,2,3,4}, reorder it to {1,4,2,3}.
+'''
+
+# Definition for singly-linked list.
+class ListNode(object):
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+
+class Solution(object):
+    def reorderList(self, head):
+        """
+        :type head: ListNode
+        :rtype: void Do not return anything, modify head in-place instead.
+        """
+        if not head:
+            return
+        # split
+        fast = slow = head
+        while fast and fast.next:
+            slow = slow.next
+            fast = fast.next.next
+        head1, head2 = head, slow.next
+        slow.next = None
+        # reverse
+        cur, pre = head2, None
+        while cur:
+            nex = cur.next
+            cur.next = pre
+            pre = cur
+            cur = nex
+        # merge
+        cur1, cur2 = head1, pre
+        while cur2:
+            nex1, nex2 = cur1.next, cur2.next
+            cur1.next = cur2
+            cur2.next = nex1
+            cur1, cur2 = nex1, nex2
+
+
+if __name__ == "__main__":
+    None

144 Binary Tree Preorder Traversal.py

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+'''
+Given a binary tree, return the preorder traversal of its nodes' values.
+
+For example:
+Given binary tree {1,#,2,3},
+   1
+    \
+     2
+    /
+   3
+return [1,2,3].
+
+Note: Recursive solution is trivial, could you do it iteratively?
+'''
+
+# Definition for a binary tree node.
+class TreeNode(object):
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+
+class Solution(object):
+    def preorderTraversal(self, root):
+        """
+        :type root: TreeNode
+        :rtype: List[int]
+        """
+        stack = []
+        result = []
+        while root or stack:
+            if not root:
+                root = stack.pop()
+            result.append(root.val)
+            if root.right:
+                stack.append(root.right)
+            root = root.left
+        return result
+
+
+if __name__ == "__main__":
+    None