LEETCODE 617. 合并二叉树
题目描述
给定两个二叉树,想象当你将它们中的一个覆盖到另一个上时,两个二叉树的一些节点便会重叠。
你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠,那么将他们的值相加作为节点合并后的新值,否则不为 NULL 的节点将直接作为新二叉树的节点。
示例 1:
输入:
Tree 1 Tree 2
1 2
/ \ / \
3 2 1 3
/ \ \
5 4 7
输出:
合并后的树:
3
/ \
4 5
/ \ \
5 4 7
注意: 合并必须从两个树的根节点开始。
代码实现
class Solution(object): def mergeTrees(self, t1, t2): """ :type t1: TreeNode :type t2: TreeNode :rtype: TreeNode """ def recursion(n1,n2): if n1 and n2: n1.val += n2.val n1.left = recursion(n1.left,n2.left) n1.right = recursion(n1.right,n2.right) return n1 return n1 or n2 return recursion(t1,t2)
另一种实现方式
class Solution(object): def mergeTrees(self, t1, t2): """ :type t1: TreeNode :type t2: TreeNode :rtype: TreeNode """ def recursion(n1,n2): root = None if n1 and n2: root = TreeNode(n1.val+n2.val) root.left = recursion(n1.left,n2.left) root.right = recursion(n1.right,n2.right) elif not n1 and n2: root = TreeNode(n2.val) root.left = recursion(None,n2.left) root.right = recursion(None,n2.right) elif n1 and not n2: root = TreeNode(n1.val) root.left = recursion(n1.left,None) root.right = recursion(n1.right,None) return root return recursion(t1,t2)7 次阅读