LeetCode 617. Merge Two Binary Trees (合并两棵二叉树)

原题

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.

Reference Answer

Method one

如果两个树都有节点的话就把两个相加，左右孩子为两者的左右孩子。

否则选不是空的节点当做子节点。

时间复杂度是O(N1+N2)，空间复杂度O(N)。N = t1 的 t2交集。

自己想尝试不建树，直接将t2值加入到t1中没能实现；而参考答案值得注意的一点事并没有进行t1 t2均为空的判定。

Code

# 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, t2):
        \"\"\"
        :type t1: TreeNode
        :type t2: TreeNode
        :rtype: TreeNode
        \"\"\"
        if not t2:
            return t1
        if not t1:
            return t2
        newT = TreeNode(t1.val + t2.val)
        newT.left = self.mergeTrees(t1.left, t2.left)
        newT.right = self.mergeTrees(t1.right, t2.right)
        return newT

Note

• 这种建立新树的过程很值得学习：

new_tree = TreeNode(t1.val + t2.val)
new_tree.left = self.mergeTrees(t1.left, t2.left)
new_tree.right = self.mergeTrees(t1.right, t2.right)

参考文献

[1] https://blog.csdn.net/fuxuemingzhu/article/details/79052953