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  • 617. Merge Two Binary Trees

    1. Quetions:  617. Merge Two Binary Trees

    url: https://leetcode.com/problems/merge-two-binary-trees/description/  

    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

    2. Solution

    class TreeNode(object):
        def __init__(self, x):
            self.val = x
            self.left = None
            self.right = None
    
    
    class Solution(object):
    
        def DFS(self, root1, root2):
    
            if root1 is None or root2 is None:
                return
    
            if root1 is not None and root2 is not None:
                root1.val += root2.val
    
            if root1.left is None:
                root1.left = root2.left
                root2.left = None
            else:
                self.DFS(root1.left, root2.left)
    
            if root1.right is None:
                root1.right = root2.right
                root2.right = None
            else:
                self.DFS(root1.right, root2.right)
    
        def mergeTrees(self, t1, t2):
            """
            :type t1: TreeNode
            :type t2: TreeNode
            :rtype: TreeNode
            """
            if t1 is None:
                return t2
    
            if t2 is None:
                return t1
    
            self.DFS(t1, t2)
    
            return t1
    

      

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  • 原文地址:https://www.cnblogs.com/ordili/p/9974868.html
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