Path Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / 
            4   8
           /   / 
          11  13  4
         /        
        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

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recursion版本

bool hasPathSum(TreeNode* root, int sum) {
    if (!root)
        return false;
    if ((root->val == sum) && (root->left == nullptr && root->right == nullptr))
        return true;
    return hasPathSum(root->left, sum - root->left->val) || hasPathSum(root->right, sum - root->right->val);
}

non-recursion版本

bool hasPathSum(TreeNode* root, int sum) {
    if (root == nullptr)
        return false;
    stack<TreeNode*> sta;
    sta.push(root);
    sum -= root->val;
    TreeNode* lastRoot = root;
    while (!sta.empty())
    {
        root = sta.top();
        if (lastRoot != root->right)
        {
            if (lastRoot != root->left) {
                if (root->left != nullptr) {
                    sta.push(root->left);
                    sum -= root->left->val;
                    continue;
                }
            }
            if (root->right != nullptr) {
                sta.push(root->right);
                sum -= root->right->val;
                continue;
            }
            else if(root->left == nullptr && sum == 0)
                return true;
        }
        lastRoot = root;
        sta.pop();
        sum += root->val;
    }
    return false;
}