import recover
import tree
if __name__ == "__main__":
sol = recover.Solution()
root = tree.construct_tree([4, 2, 6, 1, 5, 3, 7])
root.traverse_breath_first()
sol.recover(root)
root.traverse_breath_first()
from tree import TreeNode, construct_tree import verticalOrder if __name__ == "__main__": root = construct_tree([3,9,8,4,0,1,7,'#','#','#',2,5]) root.traverse_breath_first() sol = verticalOrder.Solution() print sol.verticalOrder(root)
def find_target(nums, target):
l, r = 0, len(nums) - 1
while l < r:
tmp = nums[l] + nums[r]
if tmp == target:
return True
elif tmp < target:
l += 1
else:
r -= 1
return False
if __name__ == '__main__':
root = construct_tree()
nums = inorder_traversal(root)
res = find_target(nums, 9)
print(res)
"""
一、题目
给定一个二叉搜索树和一个目标结果，如果 BST 中存在两个元素且它们的和等于给定的目标结果，则返回 true。
二、案例
输入:
5
/ \
3 6
/ \ \
2 4 7
Target = 9
输出: True（因为存在 2 + 7 = 9）
if root is None:
return root
in_order(root.left)
res.append(root.val)
in_order(root.right)
return res
def make_tree():
"""
1
\
2
/
3
"""
root = TreeNode(1)
root.right = TreeNode(2)
root.right.left = TreeNode(3)
return root
if __name__ == '__main__':
from tree import construct_tree
tree = construct_tree()
# tree = make_tree()
# print(in_order(tree))
print(in_order_traversal2(tree))
from tree import TreeNode, construct_tree
if __name__ == "__main__":
root = construct_tree(
[1, 2, 6, 3, 4, '#', '#', 7, 8, 5, '#', '#', '#', 9, '#', 10])
print root
root.traverse_pre_order()
root.traverse_breath_first()
