给定两个二叉树，想象当你将它们中的一个覆蓋到另一个上时，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。
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有很多解法能处理这个问题：主要有递归，迭代，DFS，BFS
首先想到的就是利用树的前序，中序，后序遍历处理
递归
前序遍历解法
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中序遍历解法
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后序遍历解法
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):
"""
迭代合并
: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] # 存第一个要合并的节点
q2 = [t2] # 存第二个要合并的节点
q = [t] # 存新节点
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
