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‎README.md

+5-1
Original file line numberDiff line numberDiff line change
@@ -15,5 +15,9 @@
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- [链表中倒数第k个节点](./SwordToOffer/Doc/链表中倒数第k个节点.md)
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- [反转链表](./SwordToOffer/Doc/反转链表.md)
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- [合并两个排序的链表](./SwordToOffer/Doc/合并两个排序的链表.md)
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- [树的子结构](./SwordToOffer/Doc/树的子结构.md)
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- [***树的子结构***](./SwordToOffer/Doc/树的子结构.md)
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- [***树的镜像***](./SwordToOffer/Doc/树的镜像.md)
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- [顺时针打印矩阵](./SwordToOffer/Doc/顺时针打印矩阵.md)
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‎SwordToOffer/Code/16.py

+60-2
Original file line numberDiff line numberDiff line change
@@ -5,11 +5,69 @@ def __init__(self, x):
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self.left = None
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self.right = None
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def FirstBrowse(self, pRoot):
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result = []
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if pRoot == None:
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return result
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else:
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result.append(pRoot.val)
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result = result + self.FirstBrowse(pRoot.left)
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result = result + self.FirstBrowse(pRoot.right)
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return result
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class Solution:
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def HasSubtree(self, pRoot1, pRoot2):
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# write code here
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if pRoot1==None or pRoot2==None:
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if pRoot1 == None or pRoot2 == None:
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return False
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flag = False
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if pRoot1.val == pRoot2.val:
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flag = self.doseNode1HasNode2(pRoot1, pRoot2)
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if not flag:
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flag = self.HasSubtree(pRoot1.left, pRoot2)
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if not flag:
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flag = self.HasSubtree(pRoot1.right, pRoot2)
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return flag
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def doseNode1HasNode2(self, Node1, Node2):
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if Node2 == None:
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return True
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if Node1 == None:
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return False
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if Node1.val != Node2.val:
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return False
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return self.doseNode1HasNode2(Node1.left, Node2.left) and self.doseNode1HasNode2(Node1.right, Node2.right)
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pRoot1 = TreeNode(1)
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treeNode1 = TreeNode(2)
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treeNode2 = TreeNode(3)
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treeNode3 = TreeNode(4)
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treeNode4 = TreeNode(5)
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treeNode5 = TreeNode(6)
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treeNode6 = TreeNode(7)
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pRoot1.left = treeNode1
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pRoot1.right = treeNode2
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treeNode1.left = treeNode3
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treeNode1.right = treeNode4
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treeNode2.left = treeNode5
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treeNode2.right = treeNode6
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pRoot2 = TreeNode(1)
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treeNode1b = TreeNode(2)
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treeNode2b = TreeNode(3)
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treeNode3b = TreeNode(4)
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treeNode4b = TreeNode(6)
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treeNode5b = TreeNode(7)
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pRoot2.left = treeNode1b
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pRoot2.right = treeNode2b
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treeNode2b.left = treeNode3b
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treeNode2b.right = treeNode4b
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treeNode3b.left = treeNode5b
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solution = Solution()
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print solution.HasSubtree(pRoot1, pRoot2)

‎SwordToOffer/Code/17.py

+86
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,86 @@
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# -*- coding:utf-8 -*-
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class TreeNode:
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def __init__(self, x):
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self.val = x
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self.left = None
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self.right = None
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class Solution:
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# 返回镜像树的根节点
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def Mirror(self, root):
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# write code here
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if root == None:
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return None
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if root.left == None or root.right == None:
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return self.change(root)
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if root.left.left is None and root.left.right is None and root.right.left is None and root.right.right is None:
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return self.change(root)
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else:
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mirrorRoot = TreeNode(root.val)
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mirrorRoot.left = self.Mirror(root.right)
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mirrorRoot.right = self.Mirror(root.left)
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return mirrorRoot
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def Mirror2(self, root):
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# write code here
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if not root:
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return root
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node = root.left
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root.left = root.right
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root.right = node
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self.Mirror2(root.left)
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self.Mirror2(root.right)
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return root
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def Mirror3(self,root):
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if root: # 如果存在根节点
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root.left, root.right = root.right, root.left # 根节点左右交换，俩变量交换也可以这样写
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self.Mirror(root.left) # 递归调用左节点
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self.Mirror(root.right) # 递归调用右节点
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return root # 返回节点
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else:
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return # else无节点时直接return
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def change(self, Node):
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newNode = TreeNode(Node.val)
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newNode.left = Node.right
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newNode.right = Node.left
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return newNode
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def FirstBrowse(pRoot):
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result = []
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if pRoot == None:
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return result
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else:
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result.append(pRoot.val)
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result = result + FirstBrowse(pRoot.left)
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result = result + FirstBrowse(pRoot.right)
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return result
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pRoot8 = TreeNode(8)
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treeNode6 = TreeNode(6)
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treeNode10 = TreeNode(10)
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treeNode5 = TreeNode(5)
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treeNode7 = TreeNode(7)
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treeNode9 = TreeNode(9)
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treeNode11 = TreeNode(11)
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pRoot8.left = treeNode6
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pRoot8.right = treeNode10
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treeNode6.left = treeNode5
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treeNode6.right = treeNode7
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treeNode10.left = treeNode9
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treeNode10.right = treeNode11
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s = Solution()
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t = s.Mirror(pRoot8)
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t2 = s.Mirror2(pRoot8)
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t3 = s.Mirror2(pRoot8)
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print FirstBrowse(pRoot8)
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print FirstBrowse(t)
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print FirstBrowse(t2)
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print FirstBrowse(t3)

‎SwordToOffer/Code/18.py

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Original file line numberDiff line numberDiff line change
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# -*- coding:utf-8 -*-
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import numpy as np
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class Solution:
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# matrix类型为二维列表，需要返回列表
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def printMatrix(self, matrix):
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# write code here
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size = len(matrix)
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size = int(np.sqrt(size))
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matrix = np.reshape(np.array(matrix), newshape=(size, size))
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newArray = []
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for index1 in range(0, int(size / 2)):
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for index2 in range(index1, size - index1 - 1):
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newArray.append(matrix[index1, index2])
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for index3 in range(index1, size - index1 - 1):
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newArray.append(matrix[index3, size - index1 - 1])
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for index4 in range(size - index1 - 1, index1, -1):
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newArray.append(matrix[size - index1 - 1, index4])
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for index5 in range(size - index1 - 1, index1, -1):
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newArray.append(matrix[index5, index1])
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return newArray
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def reversematrix(self,matrix):
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s=matrix.shape
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newmatrix=[]
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for i in range(s[1]-1,-1,-1):
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a=[]
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for j in range(s[0]):
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a.append(matrix[j,i])
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newmatrix.append(a)
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newmatrix=np.array(newmatrix)
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return newmatrix
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def printMatrix2(self, matrix):
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newArray=[]
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matrix=np.array(matrix)
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print matrix
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while len(matrix)!=0:
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newArray.extend(matrix[0])
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matrix=matrix[1:]
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matrix=self.reversematrix(matrix)
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return newArray
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s = Solution()
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s.printMatrix2([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]])

‎SwordToOffer/Doc/树的子结构.md

+40-1
Original file line numberDiff line numberDiff line change
@@ -6,9 +6,48 @@
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## 题目描述
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输入两棵二叉树A，B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构）
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## 解题思路
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本来的思路是采用任一一种遍历方式对两棵树同时进行遍历，然后看是否是"子串"关系，但是好像这个思路并行不通。
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正确思路应该是：
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算法1：
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先判断当前两棵树的根节点是否相同，如果相同则直接应用算法2判断pRoot2是否是pRoot1的子树，否则判断pRoot2是否是pRoot1.left的子树，如果是使用算法2，如果不是再判断pRoot2是否是pRoot1.right的子树，如果是使用算法2，如果还不是那就返回False，如果三种情况有一种满足则调用算法2并传入对应的参数。
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算法2（两个根节点相同的树，如何判断一个是否是另一个的子树）：
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首先判断Node2是否已经为None，如果是则说明Node2已经遍历完了，返回True
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如果Node2不为None则判断Node1是否为None，如果是说明Node2还没为None而Node1却为None了所以返回False
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如果Node1和Node2均不为None，递归调用算法2分别判断Node1和Node2的左右子树是否满足子树关系。
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1220
## 代码
1321

14-
[这里](../Code/16.py)
22+
```python
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def HasSubtree(self, pRoot1, pRoot2):
25+
# write code here
26+
if pRoot1 == None or pRoot2 == None:
27+
return False
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flag = False
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if pRoot1.val == pRoot2.val:
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flag = self.doseNode1HasNode2(pRoot1, pRoot2)
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if not flag:
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flag = self.HasSubtree(pRoot1.left, pRoot2)
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if not flag:
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flag = self.HasSubtree(pRoot1.right, pRoot2)
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return flag
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def doseNode1HasNode2(self, Node1, Node2):
41+
if Node2 == None:
42+
return True
43+
if Node1 == None:
44+
return False
45+
if Node1.val != Node2.val:
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return False
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return self.doseNode1HasNode2(Node1.left, Node2.left) and self.doseNode1HasNode2(Node1.right, Node2.right)
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```
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[全部代码](../Code/16.py)

‎SwordToOffer/Doc/树的镜像.md

+34
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# 树的镜像
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<center>知识点：</center>
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## 题目描述
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操作给定的二叉树，将其变换为源二叉树的镜像。
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## 输入描述
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![image-20190221212341090](https://ws3.sinaimg.cn/large/006tKfTcly1g0ed3jr4ngj30re0h83zc.jpg)
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## 解题思路
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14+
递归的思想：
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16+
```python
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def Mirror2(self, root):
18+
# write code here
19+
if not root:
20+
return root
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node = root.left
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root.left = root.right
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root.right = node
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self.Mirror2(root.left)
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self.Mirror2(root.right)
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return root
27+
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```
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## 代码
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[这里](../Code/17.py)

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