Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.
You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.
Example 1:
- Input:
Tree 1 Tree 2
1 2
/ /
3 2 1 3
/
5 4 7
- Output, Merged tree:
3
/
4 5
/
5 4 7
Note: The merging process must start from the root nodes of both trees.
C Solution:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode* mergeTrees(struct TreeNode* t1, struct TreeNode* t2) {
if (!t1 && !t2) return 0;
struct TreeNode *root = malloc(sizeof(struct TreeNode));
root->val = (t1 ? t1->val : 0) + (t2 ? t2->val : 0);
root->left = mergeTrees(t1 ? t1->left : 0, t2 ? t2->left : 0);
root->right = mergeTrees(t1 ? t1->right : 0, t2 ? t2->right : 0);
return root;
}
Python Solution 1:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def mergeTrees(self, t1, t2):
"""
:type t1: TreeNode
:type t2: TreeNode
:rtype: TreeNode
"""
if not t1 and not t2: return
if (not t1) ^ (not t2):
return t1 if t1 else t2
t1.val += t2.val
t1.left = self.mergeTrees(t1.left, t2.left)
t1.right = self.mergeTrees(t1.right, t2.right)
return t1
Python Solution 2:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def copy(self, root):
if not root: return
_root = TreeNode(root.val)
_root.left = self.copy(root.left)
_root.right = self.copy(root.right)
return _root
def mergeTrees(self, t1, t2):
"""
:type t1: TreeNode
:type t2: TreeNode
:rtype: TreeNode
"""
if not t1 and not t2: return
if (not t1) ^ (not t2):
return self.copy(t1 if t1 else t2)
root = TreeNode(t1.val + t2.val)
root.left = self.mergeTrees(t1.left, t2.left)
root.right = self.mergeTrees(t1.right, t2.right)
return root
Summary:
- Could I modify the t1 or t2?
- If yes, free the abandoned one.
LeetCode: 617. Merge Two Binary Trees
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