# 513 Find Bottom Left Tree Value

Given a binary tree, find the leftmost value in the last row of the tree.

**Example 1:**

```
Input:

    2
   / \
  1   3

Output:
1
```

**Example 2:**

```
Input:

        1
       / \
      2   3
     /   / \
    4   5   6
       /
      7

Output:
7
```

**Note:**You may assume the tree (i.e., the given root node) is not **NULL**.

**Complexity**: O(n) time, O(logn) space

**The Idea**: Run a level order traversal with a small twist: check for instances when the next level arises (left to right top to bottom), and collect the first left instance.

```cpp
int findBottomLeftValue(TreeNode* root) {
    if (!root) return -1;

    int deepest_left = root->val;
    queue<TreeNode*> q;
    q.push(root);
    q.push(nullptr);

    while (!q.empty()) {
        TreeNode *temp = q.front();
        q.pop();

        if (temp) {
            if (temp->left)
                q.push(temp->left);
            if (temp->right)
                q.push(temp->right);
        }
        else {
            if (!q.empty() && q.front()) {
                deepest_left = q.front()->val;
                q.push(nullptr);
            }
        }
    }

    return deepest_left;
}
```


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