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。
解题思路:
二叉树问题,使用较多的办法是递归
C++
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) { if (t1 == NULL) return t2; if (t2 == NULL) return t1; TreeNode* ret = new TreeNode(t1->val + t2->val); ret->left = mergeTrees(t1->left, t2->left); ret->right = mergeTrees(t1->right, t2->right); return ret; } };
Python
# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def mergeTrees(self, t1: TreeNode, t2: TreeNode) -> TreeNode: if t1 is None: return t2 if t2 is None: return t1 ret = TreeNode(t1.val + t2.val) ret.left = self.mergeTrees(t1.left, t2.left) ret.right = self.mergeTrees(t1.right, t2.right) return ret