[Leetcode] Binary tree -- 617. Merge Two Binary Trees

Posted 2020-10-01 安新

 

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.

 

Solution:

 

1. use recursive time complexity o(min(m,n)) m is vertex number of t1, n is the vertex number of t2

#combine the node to left tree t1. If node are overlapped, then add valu to t1.val. else return t1 or t2 if they are null repsectively.

1         if t1 is None:
2             return t2
3         if t2 is None:
4             return t1
5         t1.val += t2.val
6         t1.left = mergeTrees(t1.left, t2.left)
7         t1.right = mergeTrees(t1.right, t2.right)
8         
9         return t1

 

2.   use iterative way

 1        if t1 is None:
 2             return t2
 3         st = []
 4         st.append((t1,t2))
 5         while (len(st)):
 6             nodes = st.pop()
 7             r1 = nodes[0]
 8             r2 = nodes[1]
 9             if r1 is None or r2 is None:
10                 continue
11             r1.val += r2.val
12             st.append((r1.left, r2.left))
13             st.append((r1.right, r2.right))
14                 
15             if r1.left is None and r2.left is not None:
16                 r1.left = r2.left
17             if r1.right is None and r2.right is not None:
18                 r1.right = r2.right
19 
20         return t1