404. Sum of Left Leaves 难度:easy
2016-10-13 21:35
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题目:
Find the sum of all left leaves in a given binary tree.
Example:
思路:
要求我们计算书中所有左叶子节点权值的和。根据节点的类型,分别用不同方式递归地调用函数,当节点为NULL则返回0,当节点有左孩子且左孩子为叶子节点,则返回左孩子的权值加上以右孩子为根的树中左叶子节点权值的和,若左孩子不是叶子节点,则返回分别以左右孩子为根的树的左孩子节点权值和的和,如果没有左孩子节点,则返回以右孩子为根的树中左叶子节点权值的和。
程序:
/**
* 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(root == NULL)
return 0;
else if(root->left != NULL)
{
if(root->left->left == NULL&&root->left->right == NULL)
return root->left->val + sumOfLeftLeaves(root->right);
else
return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
}
else
return sumOfLeftLeaves(root->right);
}
};
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.
思路:
要求我们计算书中所有左叶子节点权值的和。根据节点的类型,分别用不同方式递归地调用函数,当节点为NULL则返回0,当节点有左孩子且左孩子为叶子节点,则返回左孩子的权值加上以右孩子为根的树中左叶子节点权值的和,若左孩子不是叶子节点,则返回分别以左右孩子为根的树的左孩子节点权值和的和,如果没有左孩子节点,则返回以右孩子为根的树中左叶子节点权值的和。
程序:
/**
* 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(root == NULL)
return 0;
else if(root->left != NULL)
{
if(root->left->left == NULL&&root->left->right == NULL)
return root->left->val + sumOfLeftLeaves(root->right);
else
return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
}
else
return sumOfLeftLeaves(root->right);
}
};
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