LeetCode #617 合并二叉树 树 递归

LeetCode #617 合并二叉树

题目描述

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。

方法一：递归

# 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 and t2:
            return t2
        elif t1 and t2:
            t1.val = t1.val + t2.val
            t1.left = self.mergeTrees(t1.left, t2.left)
            t1.right = self.mergeTrees(t1.right, t2.right)
        # t1 and not t2
        # else:
        #     return t1
        return t1

• 时间复杂度： $O(N)$
• 空间复杂度： $O(N)$

方法二：递归

# 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 and t2:
            return t2
        elif t1 and t2:
            stack = [(t1, t2)]
            while stack:
                r1, r2 = stack.pop(0)
                r1.val += r2.val
                # 如果r1和r2的左子树都存在就把r2的左子树挂在r1上
                if r1.left and r2.left:
                    stack.append((r1.left, r2.left))
                elif not r1.left:
                    r1.left = r2.left
                # 右子树也是一样的
                if r1.right and r2.right:
                    stack.append((r1.right, r2.right))
                elif not r1.right:
                    r1.right = r2.right
        # t1 and not t2
        # else:
        #     return t1
        return t1

• 时间复杂度： $O(N)$
• 空间复杂度： $O(N)$