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.
class Solution {
public:
int sumOfLeftLeaves(TreeNode* root) {
if (!root) return 0;
int sum = 0;
if (root->left!=NULL)
{
TreeNode * lroot = root->left;
if(lroot->left==NULL&&lroot->right==NULL)
{
sum += lroot->val;
}
else
{
sum += sumOfLeftLeaves(lroot);
}
}
sum+=sumOfLeftLeaves(root->right);
return sum;
}
};
本文介绍了一种算法,用于求解给定二叉树中所有左叶子节点的值之和。通过递归遍历的方式,判断每个节点是否为左叶子节点并累加其值,最终返回总和。
781

被折叠的 条评论
为什么被折叠?



