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404. Sum of Left Leaves

Description

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

A leaf is a node with no children. A left leaf is a leaf that is the left child of another node.

 

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

 

Constraints:

  • The number of nodes in the tree is in the range [1, 1000].
  • -1000 <= Node.val <= 1000

Solutions

Solution 1: Recursion

First, we check if root is null. If it is, we return $0$.

Otherwise, we recursively call the sumOfLeftLeaves function to calculate the sum of all left leaves in root's right subtree, and assign the result to the answer variable $ans$. Then we check if root's left child exists. If it does, we check if it is a leaf node. If it is a leaf node, we add its value to the answer variable $ans$. Otherwise, we recursively call the sumOfLeftLeaves function to calculate the sum of all left leaves in root's left subtree, and add the result to the answer variable $ans$.

Finally, we return the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of nodes in the binary tree.

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:
        if root is None:
            return 0
        ans = self.sumOfLeftLeaves(root.right)
        if root.left:
            if root.left.left == root.left.right:
                ans += root.left.val
            else:
                ans += self.sumOfLeftLeaves(root.left)
        return ans

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int ans = sumOfLeftLeaves(root.right);
        if (root.left != null) {
            if (root.left.left == root.left.right) {
                ans += root.left.val;
            } else {
                ans += sumOfLeftLeaves(root.left);
            }
        }
        return ans;
    }
}

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if (!root) {
            return 0;
        }
        int ans = sumOfLeftLeaves(root->right);
        if (root->left) {
            if (!root->left->left && !root->left->right) {
                ans += root->left->val;
            } else {
                ans += sumOfLeftLeaves(root->left);
            }
        }
        return ans;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func sumOfLeftLeaves(root *TreeNode) int {
    if root == nil {
        return 0
    }
    ans := sumOfLeftLeaves(root.Right)
    if root.Left != nil {
        if root.Left.Left == root.Left.Right {
            ans += root.Left.Val
        } else {
            ans += sumOfLeftLeaves(root.Left)
        }
    }
    return ans
}

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/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function sumOfLeftLeaves(root: TreeNode | null): number {
    if (!root) {