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.

題意很簡單，直接做DFS深度優先遍歷即可，在遍歷的時候作比較即可

程式碼如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
 

#include <cmath>

using namespace std;


/*
struct TreeNode 
{
     int val;
     TreeNode *left;
     TreeNode *right;
     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
*/

class Solution 
{
public:
    int findBottomLeftValue(TreeNode* root) 
    {
        if (root == NULL)
            return
 
 0;

        int maxDepth = 0 , res = root->val;
        getAll(root,maxDepth,0,res);
        return res;
    }

    void getAll(TreeNode* root, int& maxDepth, int depth,int& res)
    {
        if (root == NULL)
            return;
        else
        {
            getAll(root->left,  maxDepth, depth + 1, res);
            getAll(root->right, maxDepth, depth + 1, res);
            if (depth > maxDepth)
            {
                maxDepth = depth;
                res = root->val;
            }
            return;
        }
    }
};