【leetcode】617. Merge Two Binary Trees

题目如下：

解题思路：非常简单的一个题目，创建新的树根节点newNode，然后先遍历t1，把t1复制一遍给newNode；接下来遍历t2,如果t2和newNode两者对应的相同位置的节点都存在，那么newNode对应位置的节点的值加上t2对应的值；如果newNode不存在，则创建一个和t2对应位置的节点值一样的新节点。

代码如下：

# 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 traverse(self,newNode,node):
        if node.left != None:
            if newNode.left == None:
                newNode.left = TreeNode(node.left.val)
            else:
                newNode.left.val += node.left.val
            self.traverse(newNode.left,node.left)
        if node.right != None:
            if newNode.right == None:
                newNode.right = TreeNode(node.right.val)
            else:
                newNode.right.val += node.right.val
            self.traverse(newNode.right, node.right)


    def mergeTrees(self, t1, t2):
        """
        :type t1: TreeNode
        :type t2: TreeNode
        :rtype: TreeNode
        """
        root = None
        if t1 == t2 == None:
            return None
        if t1 != None:
            root = TreeNode(t1.val)
            self.traverse(root,t1)
        if t2 != None:
            if root == None:
                root = TreeNode(t2.val)
            else:
                root.val += t2.val
            self.traverse(root, t2)
        return root
        

转载于:https://www.cnblogs.com/seyjs/p/9660715.html