算法练习之平衡二叉树

Posted 2021-04-02 shelly笔记

1. 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 = x self.left = None self.right = Noneclass Solution: def test(self, tree1: TreeNode, tree2: TreeNode): p = [] def search(tree1, tree2, x, y): if not tree1 and not tree2: return if not tree1 or not tree2: p.append((x, y)) return print(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