construct binary tree from preorder and postorder traversal medium contest

Author: weiy

Tree/ConstructBinaryTreeFromPreorderAndPostorderTraversal.py

@@ -0,0 +1,128 @@
+"""
+Return any binary tree that matches the given preorder and postorder traversals.
+
+Values in the traversals pre and post are distinct positive integers.
+
+ 
+
+Example 1:
+
+Input: pre = [1,2,4,5,3,6,7], post = [4,5,2,6,7,3,1]
+Output: [1,2,3,4,5,6,7]
+ 
+
+Note:
+
+1 <= pre.length == post.length <= 30
+pre[] and post[] are both permutations of 1, 2, ..., pre.length.
+It is guaranteed an answer exists. If there exists multiple answers, you can return any of them.
+
+根据二叉树的 前序和后序遍历，返回一颗完整的二叉树。
+
+不唯一，返回随便一个即可。
+
+思路：
+1. 二叉树的前序是 根左右。
+2. 二叉树的后序是 左右根。
+
+在前序中确定 根 ，然后在后序中找左右子树。
+
+pre = [1,2,4,5,3,6,7]
+post = [4,5,2,6,7,3,1]
+
+总根是 1
+
+1的左右其中一个是 2，就当它是左子树好了，因为是 根 左右 所以假设的话就先假设为左子树，如果只有一边的话，左右其实无所谓。
+
+   1
+ /
+2
+
+然后在 post 中找属于 2 这个子树的节点。
+找到 4 5 2，那么剩下的 6 7 3 就是与之相对的 1 的右子树了。
+
+把 pre 分成 [2, 4, 5] [3, 6, 7]
+   post 分为 [4, 5, 2] [6, 7 ,3]
+
+这样 作为1的左右两颗子树已经出来了。
+pre[left] 和 pre[right] 的 0 分别为 左右两棵子树的根。
+
+之后就是分别把左右两边的这两个代替原来的 pre post，根也同样代替，然后递归直到没有即可。
+
+测试链接：
+https://leetcode.com/contest/weekly-contest-98/problems/construct-binary-tree-from-preorder-and-postorder-traversal/
+
+beat 50% 40ms.
+
+这应该是自己的极限了。
+4道题 一个半小时做 3 道题，1 easy 2 medium 0 hard.
+
+
+"""
+
+# 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 constructFromPrePost(self, pre, post):
+        """
+        :type pre: List[int]
+        :type post: List[int]
+        :rtype: TreeNode
+        """
+        
+        def getLeftAndRight(pre, post):            
+            # no more node.
+            if not pre:
+                return None
+            
+            # Get the index of the left root.
+            index = post.index(pre[0])
+            
+            # post left tree
+            val_children = post[:index+1]
+            
+            # post right tree
+            val_brother = post[index+1:-1]
+            
+            # pre left tree
+            # Get the left tree pre list 
+            # if left tree post list contains 3 elements
+            # then we will get equal in pre list.
+            t = len(val_children)
+            pre_val_children = pre[:t]
+            
+            # there is right tree.
+            # The elements are the rest of pre list.
+            pre_val_brother = pre[t:]
+
+            left = pre[0]
+            
+            right = val_brother[-1] if val_brother else None
+
+            # left, right, pre, post
+            return (left, right, pre_val_children, val_children, pre_val_brother, val_brother)
+            
+        def construct(root, pre, post):
+            x = getLeftAndRight(pre[1:], post)
+            if root and x:
+                if x[0] is not None:
+                    root.left = TreeNode(x[0])
+                if x[1] is not None:
+                    root.right = TreeNode(x[1])
+
+                if root.left:
+                    construct(root.left, x[2], x[3])
+
+                if root.right:
+                    construct(root.right, x[4], x[5])
+            
+        allRoot = TreeNode(pre[0])
+        construct(allRoot, pre, post)
+        
+        return allRoot
+