Find the sum of all left leaves in a given binary tree.
Example:
3
/ 9 20
/ 15 7
There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.
计算给定二叉树的所有左叶子之和。
示例：
3
/ 9 20
/ 15 7
在这个二叉树中，有两个左叶子，分别是 9 和 15，所以返回 24
1 /**
2 * Definition for a binary tree node.
3 * public class TreeNode {
4 * public var val: Int
5 * public var left: TreeNode?
6 * public var right: TreeNode?
7 * public init(_ val: Int) {
8 * self.val = val
9 * self.left = nil
10 * self.right = nil
11 * }
12 * }
13 */
14 class Solution {
15 var sum:Int = 0
16 func sumOfLeftLeaves(_ root: TreeNode?) -> Int {
17 if root == nil {return sum}
18 if root!.left != nil
19 {
20 if root!.left!.left == nil && root!.left!.right == nil
21 {
22 sum += root!.left!.val
23 }
24 }
25 //递归
26 sumOfLeftLeaves(root!.left)
27 sumOfLeftLeaves(root!.right)
28 return sum
29 }
30 }
12ms
1 /**
2 * Definition for a binary tree node.
3 * public class TreeNode {
4 * public var val: Int
5 * public var left: TreeNode?
6 * public var right: TreeNode?
7 * public init(_ val: Int) {
8 * self.val = val
9 * self.left = nil
10 * self.right = nil
11 * }
12 * }
13 */
14 class Solution {
15 func sumOfLeftLeaves(_ root: TreeNode?) -> Int {
16 return sumOfLeftLeaves(root, false)
17 }
18
19 func sumOfLeftLeaves(_ node: TreeNode?, _ isLeft: Bool) -> Int {
20 guard let node = node else { return 0 }
21 if node.left == nil && node.right == nil && isLeft {
22 return node.val
23 }
24
25 return sumOfLeftLeaves(node.left, true) + sumOfLeftLeaves(node.right, false)
26 }
27 }
16ms
1 /**
2 * Definition for a binary tree node.
3 * public class TreeNode {
4 * public var val: Int
5 * public var left: TreeNode?
6 * public var right: TreeNode?
7 * public init(_ val: Int) {
8 * self.val = val
9 * self.left = nil
10 * self.right = nil
11 * }
12 * }
13 */
14 class Solution {
15 var sum = 0
16
17 func sumOfLeftLeaves(_ root: TreeNode?) -> Int {
18 if root == nil || (root?.left == nil && root?.right == nil) {
19 return 0
20 }
21
22 search(root)
23
24 return sum
25 }
26
27 func search(_ root: TreeNode?) -> Void {
28 if root == nil {
29 return
30 }
31
32 if isLeave(root?.left) {
33 let val = root?.left?.val ?? 0
34 sum = sum + val
35 } else {
36 search(root?.left)
37 }
38
39 if isLeave(root?.right) {
40 return
41 } else {
42 search(root?.right)
43 }
44 }
45
46 func isLeave(_ root: TreeNode?) -> Bool {
47 if root == nil {
48 return true
49 }
50
51 if root?.left == nil && root?.right == nil {
52 return true
53 }
54
55 return false
56 }
57 }
16ms
1 /**
2 * Definition for a binary tree node.
3 * public class TreeNode {
4 * public var val: Int
5 * public var left: TreeNode?
6 * public var right: TreeNode?
7 * public init(_ val: Int) {
8 * self.val = val
9 * self.left = nil
10 * self.right = nil
11 * }
12 * }
13 */
14 class Solution {
15 func sumOfLeftLeaves(_ root: TreeNode?) -> Int {
16 var result = 0
17 sumOfLeftLeaves(root, &result)
18 return result
19 }
20
21 func sumOfLeftLeaves(_ root: TreeNode?, _ sum: inout Int) {
22 guard let root = root else {
23 return
24 }
25
26 if let left = root.left {
27 if left.left == nil && left.right == nil {
28 sum += left.val
29 } else {
30 sumOfLeftLeaves(left, &sum)
31 }
32 }
33
34 sumOfLeftLeaves(root.right, &sum)
35 }
36 }
24ms
1 /**
2 * Definition for a binary tree node.
3 * public class TreeNode {
4 * public var val: Int
5 * public var left: TreeNode?
6 * public var right: TreeNode?
7 * public init(_ val: Int) {
8 * self.val = val
9 * self.left = nil
10 * self.right = nil
11 * }
12 * }
13 */
14 class Solution {
15 func sumOfLeftLeaves(_ root: TreeNode?) -> Int {
16 var sum = 0
17 toSumLeftTreeNode(root: root, sum: &sum)
18 return sum
19 }
20 func toSumLeftTreeNode(root: TreeNode?, sum: inout Int){
21 if root == nil {
22 return
23 }
24 if root?.left != nil && (root?.left?.left == nil && root?.left?.right == nil){
25 sum = sum + (root?.left?.val)!
26 }
27 toSumLeftTreeNode(root: root?.left, sum: &sum)
28 toSumLeftTreeNode(root: root?.right, sum: &sum)
29 }
30 }
原文地址：https://www.cnblogs.com/strengthen/p/9782487.html
时间： 2024-11-02 12:25:12