原題

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

如果兩個樹都有節點的話就把兩個相加，左右孩子為兩者的左右孩子。

否則選不是空的節點當做子節點。

時間複雜度是O(N1+N2)，空間複雜度O(N)。N = t1 的 t2交集。

自己想嘗試不建樹，直接將t2值加入到t1中沒能實現；而參考答案值得注意的一點事並沒有進行t1 t2均為空的判定。

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

• 這種建立新樹的過程很值得學習：

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)

參考文獻

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