https://leetcode.com/problems/sum-of-left-leaves/description/
题目
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.
题目翻译
题目解析
参考答案
#include <iostream>
using namespace std;
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
bool bIsLeaves(TreeNode * node)
{
if (NULL == node)
{
return false;
}
if (node->left == NULL && node->right == NULL)
{
return true;
}
else
{
return false;
}
}
int sumOfLeftLeavesRecur(TreeNode* root) {
if (NULL == root)
{
return 0;
}
if (bIsLeaves(root))
{
return root->val;
}
int iLeftSum = sumOfLeftLeavesRecur(root->left);
int iRightSum = sumOfLeftLeavesRecur(root->right);
if (bIsLeaves(root->right))
{
iRightSum -= root->right->val;
}
return iLeftSum + iRightSum;
}
int sumOfLeftLeaves(TreeNode* root) {
if (NULL == root)
{
return 0;
}
if (bIsLeaves(root))
{
return 0;
}
return sumOfLeftLeavesRecur(root);
}
};
int main()
{
return 0;
}
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int sumOfLeftLeaves(TreeNode* root) {
if(NULL == root)
{
return 0;
}
int iSum = 0;
if(root->left != NULL && (root->left->left == NULL && root->left->right == NULL))
{
iSum = root->left->val;
}
return iSum + sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
}
};
