算法练习之平衡二叉树
2棵二叉平衡树，找出值不相等的节点，打印节点坐标（位置信息）
0
/ \
3 5
/ \
7 10
2
/ \
6 9
/ \
7 8
思路：明显的递归遍历（深度遍历或广度遍历）这里就用比较简单的方法，深度遍历了。
思考了很久如何获取坐标位置，本来是用闭包来做，但是好像修改后获取到的坐标不准确了，改为参数传入了，一起递归。
# Definition for a binary tree node.class TreeNode:def __init__(self, x):self.val = xself.left = Noneself.right = Noneclass Solution:def test(self, tree1: TreeNode, tree2: TreeNode):p = []def search(tree1, tree2, x, y):if not tree1 and not tree2:returnif not tree1 or not tree2:p.append((x, y))returnprint(x, y, tree1.val, tree2.val)if tree1.val != tree2.val:p.append((x, y))search(tree1.left, tree2.left, x*2, y+1)search(tree1.right, tree2.right, x*2+1, y+1)search(tree1, tree2, 0, 0)return pif __name__ == "__main__":sl = Solution()t1 = TreeNode(0)t1.left = TreeNode(4)t1.right = TreeNode(5)t1.left.left = TreeNode(6)t1.left.right = TreeNode(7)t2 = TreeNode(2)t2.left = TreeNode(2)t2.right = TreeNode(9)r = sl.test(t1, t2)print(r)
做完了发现其实很容易，但是做了一个小时诶，平时做得比较多的是动态规划类的题目，碰到树的题目还是有点害怕的，还是要多练习多练习