Algorithem_Merge Two Binary Trees
原创Algorithem_Merge Two Binary Trees
原创
You are given two binary trees root1 and root2.
Imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge the two trees into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of the new tree.
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Return the merged tree.
Note: The merging process must start from the root nodes of both trees.
Example 1:
Input: root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]
Output: [3,4,5,5,4,null,7]Example 2:
Input: root1 = [1], root2 = [1,2]
Output: [2,2]解法
使用递归，看上面的例子能看出，基础计算是两个值相加，如果有一个为 null，则和0相加。
代码逻辑：
声明一个新的 TreeNode，TreeNode 的val = root1.val + root2.val，TreeNode 的 left=递归(root1.left, roo2.left)，right=递归(root1.right, root2.right)
代码如下：
/**
* Definition for a binary tree node.
* public class TreeNode {
* public var val: Int
* public var left: TreeNode?
* public var right: TreeNode?
* public init() { self.val = 0; self.left = nil; self.right = nil; }
* public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
* public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
* self.val = val
* self.left = left
* self.right = right
* }
* }
*/
class Solution {
func mergeTrees(_ root1: TreeNode?, _ root2: TreeNode?) -> TreeNode? {
if root1 == nil && root2 == nil {
return nil
}
let valValue = (root1 == nil ? 0 : root1!.val) + (root2 == nil ? 0 : root2!.val)
let newNode = TreeNode(valValue)
newNode.left = mergeTrees(root1?.left, root2?.left)
newNode.right = mergeTrees(root1?.right, root2?.right)
return newNode
}
}原创声明：本文系作者授权腾讯云开发者社区发表，未经许可，不得转载。
如有侵权，请联系 cloudcommunity@tencent.com 删除。
原创声明：本文系作者授权腾讯云开发者社区发表，未经许可，不得转载。
如有侵权，请联系 cloudcommunity@tencent.com 删除。