【剑指Offer学习】【面试题25：二叉树中和为某一值的路径】

 

https://blog.youkuaiyun.com/feit2417/article/details/99544525

typedef struct treeNode
{
    int m_nValue;
    struct treeNode* left;
    struct treeNode* right;
} tree;

bool doFindPath(tree* root, int needNum, vector<int> &path) {
    path.push(root->m_nValue);
    int rootVlue = root->m_nValue;
    if (rootVlue == needNum) {
        // g如果需要的数值和当前数组相等且是叶子结点，那么打印路径
        if (root->left==NULL && root->right==NULL) {
            printPath(path);
        }
    }
    if (root->left) {
        doFindPath(root->left,needNum-rootVlue,path);
    }
    if (root->right) {
        doFindPath(root->right,needNum-rootVlue,path);
    }
    //递归回溯退回父节点，在路径上要删除当前节点
    path.pop();
}
bool findPath(tree* root, int num) {
    if (root == NULL) {
        return false;
    }
    vector<int> path;
    doFindPath(root,num,path);
}