leetcode 404 Sum of Left Leaves

Findthesumofallleftleavesinagivenbinarytree.

Example:

3
/\
920
/\
157

Therearetwoleftleavesinthebinarytree,withvalues9and15respectively.Return24.

分析：关键是怎么判断它是左叶子；

/**
*Definitionforabinarytreenode.
*structTreeNode{
*intval;
*TreeNode*left;
*TreeNode*right;
*TreeNode(intx):val(x),left(NULL),right(NULL){}
*};
*/
classSolution{
public:
intsumOfLeftLeaves(TreeNode*root){
if(root==NULL)
return0;
TreeNode*temp=root->left;
if(temp&&(temp->left==NULL)&&(temp->right==NULL))
returntemp->val+sumOfLeftLeaves(root->right);
else
returnsumOfLeftLeaves(root->left)+sumOfLeftLeaves(root->right);
}
};

也可用bfs

网上大神的dfs：深度优先遍历，将所有结点从根结点开始遍历一遍，设立isLeft的值，当当前结点是叶子节点并且也是左边，那就result加上它的值

/**
*Definitionforabinarytreenode.
*structTreeNode{
*intval;
*TreeNode*left;
*TreeNode*right;
*TreeNode(intx):val(x),left(NULL),right(NULL){}
*};
*/
classSolution{
public:
intresult=0;
intsumOfLeftLeaves(TreeNode*root){
if(root==NULL)
return0;
dfs(root,false);
returnresult;
}
voiddfs(TreeNode*root,boolisLeft){
if(root->left==NULL&&root->right==NULL){
if(isLeft==true)
result+=root->val;
return;
}
if(root->left!=NULL)
dfs(root->left,true);
if(root->right!=NULL)
dfs(root->right,false);
}
};