Find Bottom Left Tree Value

Description: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.
思路：该题目是找二叉树的最左边的 叶子结点，首先想到的是用BFS，对二叉树进行层序遍历，最关键的是要记录当前二叉树的层级，又利用了一个队列记录了每个结点的层级
/**
* 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 findBottomLeftValue(TreeNode* root) {
queue<TreeNode*> q;
queue<int> level;
int l = 1;
q.push(root);
level.push(l);
4000
TreeNode* tmp = root;
int ret = root->val;
while(!q.empty())
{
TreeNode* t = q.front();
int ll = level.front();

q.pop();
level.pop();

if(l != ll)
{
l = ll;
ret = t->val;
}

ll = ll + 1;
if(t->left != NULL)
{
q.push(t->left);
level.push(ll);
}

if(t->right != NULL)
{
q.push(t->right);
level.push(ll);
}
}

return ret;
}
};