合併兩個二叉樹

給定兩個二叉樹，想象當你將它們中的一個覆蓋到另外一個上時，兩個二叉樹的一些節點便會重疊。

你須要將他們合併爲一個新的二叉樹。合併的規則是若是兩個節點重疊，那麼將他們的值相加做爲節點合併後的新值，不然不爲 NULL 的節點將直接做爲新二叉樹的節點。

輸入: 
	Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  
輸出: 
合併後的樹:
	     3
	    / \
	   4   5
	  / \   \ 
	 5   4   7

有不少解法能處理這個問題：主要有遞歸，迭代，DFS，BFS

首先想到的就是利用樹的前序，中序，後序遍歷處理

遞歸

前序遍歷解法

class Solution(object):
    def mergeTrees(self, t1, t2):
        if t1 is None and t2 is None:
            return
        if t1 is None:
            return t2
        if t2 is None:
            return t1
        t = TreeNode(t1.val + t2.val)
        t.left = self.mergeTrees(t1.left, t2.left)
        t.right = self.mergeTrees(t1.right, t2.right)
        return t

中序遍歷解法

class Solution(object):
    def mergeTrees(self, t1, t2):
        if t1 is None and t2 is None:
            return
        if t1 is None:
            return t2
        if t2 is None:
            return t1
        t_left = self.mergeTrees(t1.left, t2.left)
        t = TreeNode(t1.val + t2.val)
        t.left = t_left
        t.right = self.mergeTrees(t1.right, t2.right)
        return t

後序遍歷解法

class Solution(object):
    def mergeTrees(self, t1, t2):
        if t1 is None and t2 is None:
            return
        if t1 is None:
            return t2
        if t2 is None:
            return t1
        t_left = self.mergeTrees(t1.left, t2.left)
        t_right = self.mergeTrees(t1.right, t2.right)
        t = TreeNode(t1.val + t2.val)
        t.left = t_left
        t_right = t_right
        return t

迭代，BFS

class Solution(object):
    def mergeTrees(self, t1, t2):
        """
        迭代合併
        :param t1:
        :param t2:
        :return:
        """
        if t1 is None and t2 is None:
            return None
        if t1 is None:
            return t2
        if t2 is None:
            return t1
        t = TreeNode(t1.val + t2.val)
        q1 = [t1]  # 存第一個要合併的節點
        q2 = [t2]  # 存第二個要合併的節點
        q = [t]  # 存新節點
        while len(q) > 0:
            node = q.pop(0)
            t1 = q1.pop(0)
            t2 = q2.pop(0)
            
            if t1.left is None and t2.left is None:
                pass
            elif t1.left is None:
                node.left = t2.left
            elif t2.left is None:
                node.left = t1.left
            else:
                node.left = TreeNode(t1.left.val + t2.left.val)
                q.append(node.left)
                q1.append(t1.left)
                q2.append(t2.left)
                
            if t1.right is None and t2.right is None:
                pass
            elif t1.right is None:
                node.right = t2.right
            elif t2.right is None:
                node.right = t1.right
            else:
                node.right = TreeNode(t1.right.val + t2.right.val)
                q.append(node.right)
                q1.append(t1.right)
                q2.append(t2.right)
                
        return t