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# 404. Sum of Left Leaves

## 題目 :

Given the root of a binary tree, return the sum of all left leaves.

### Example :

Note

Example 1:

Input: root = [3,9,20,null,null,15,7] Output: 24 Explanation: There are two left leaves in the binary tree, with values 9 and 15 respectively.

Example 2:

Input: root = [1] Output: 0

#### 解題思路 :

• 這題是用recursive的方式實現DFS的概念(大部分有關於leaves的題目都可以往DFS的方向去思考)
• 題目需要把每一個left leaves加起來，判斷一個leave是否為left leaves就是他是否是上一個node的left children
• 這裡提供兩個解法，兩個解法大同小異，一個是straight forward的去寫出邏輯判斷node.Left 跟 node.Left.Left 跟 node.Left.Right，另一個是在argument中加入一個boolean去判斷上一個該node是否是left children

### Recursive解法-1

• time complexity: O(n) , space complexity: O(1)

Runtime: 0 ms, faster than 100.00% of Go online submissions for Sum of Left Leaves.

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 `````` ``````/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ func sumOfLeftLeaves(root *TreeNode) int { var sum int findLeftLeaves(root, &sum) return sum } func findLeftLeaves(node *TreeNode, sum *int) { if node.Left != nil && node.Left.Left == nil && node.Left.Right == nil { *sum += node.Left.Val } if node.Left != nil { findLeftLeaves(node.Left, sum) } if node.Right != nil { findLeftLeaves(node.Right, sum) } } ``````

### Recursive解法-2

• time complexity: O(n) , space complexity: O(1)

Runtime: 0 ms, faster than 100.00% of Go online submissions for Sum of Left Leaves.

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 `````` ``````/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ func sumOfLeftLeaves(root *TreeNode) int { return getLeftLeavesSum(root, false) } func getLeftLeavesSum(node *TreeNode, isLeft bool) int { if node == nil { return 0 } if node.Left == nil && node.Right == nil { if isLeft { return node.Val }else { return 0 } } leftSum := getLeftLeavesSum(node.Left, true) rightSum := getLeftLeavesSum(node.Right, false) return leftSum + rightSum } ``````