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[Leetcode]404. Sum of Left Leaves (C++)

题目描述

题目链接:404. Sum of Left Leaves

Find the sum of all left leaves in a given binary tree.

例子

3
/ \
9 20
/ \
15 7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

解题思路

这道题思路比较简单,用各种树的遍历方法都可以,只需要给每一个节点加一个是否左节点的标识即可,遇到叶节点时同时是左节点的情况下就加入到结果中,代码如下:

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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) {
int sum = 0;
dfs(root, false, sum);
return sum;
}

private:
void dfs(TreeNode* node, bool is_left, int& sum) {
if (!node) return;
if (!node->left && !node->right) sum += is_left ? node->val : 0;
dfs(node->left, true, sum);
dfs(node->right, false, sum);
}
};
  • 时间复杂度:O(n)
  • 空间复杂度:O(k) k 为树最大深度