617. 合并二叉树

给定两个二叉树，想象当你将它们中的一个覆盖到另一个上时，两个二叉树的一些节点便会重叠。

你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠，那么将他们的值相加作为节点合并后的新值，否则不为 NULL 的节点将直接作为新二叉树的节点。

示例 1:

输入: 
	Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  
输出: 
合并后的树:
	     3
	    / \
	   4   5
	  / \   \ 
	 5   4   7

注意: 合并必须从两个树的根节点开始。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/merge-two-binary-trees

思路及算法

深度优先搜索

同时对两个树进行【深度优先搜索】

一共三种情况

1. t1节点和t2节点均为null，合并之后的新节点为null
2. t1与t2节点两者有一个为null，合并之后的新节点为两者中的非空节点
3. t1与t2节点均不为null，合并则为两者相加后的新节点

# 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: TreeNode, t2: TreeNode) -> TreeNode:
        if not t1:
            return t2
        if not t2:
            return t1
        
        r = TreeNode(t1.val + t2.val)
        r.left = self.mergeTrees(t1.left, t2.left)
        r.right = self.mergeTrees(t1.right, t2.right)
        return r

广度优先搜索

# 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: TreeNode, t2: TreeNode) -> TreeNode:
        if not t1:
            return t2
        if not t2:
            return t1
        
        merged = TreeNode(t1.val + t2.val)
        queue = collections.deque([merged])
        queue1 = collections.deque([t1])
        queue2 = collections.deque([t2])

        while queue1 and queue2:
            node = queue.popleft()
            node1 = queue1.popleft()
            node2 = queue2.popleft()
            left1, right1 = node1.left, node1.right
            left2, right2 = node2.left, node2.right
            if left1 or left2:
                if left1 and left2:
                    left = TreeNode(left1.val + left2.val)
                    node.left = left
                    queue.append(left)
                    queue1.append(left1)
                    queue2.append(left2)
                elif left1:
                    node.left = left1
                elif left2:
                    node.left = left2
            if right1 or right2:
                if right1 and right2:
                    right = TreeNode(right1.val + right2.val)
                    node.right = right
                    queue.append(right)
                    queue1.append(right1)
                    queue2.append(right2)
                elif right1:
                    node.right = right1
                elif right2:
                    node.right = right2
        
        return merged