题目描述：

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

Example 1:

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Input:

    2
   / \
  1   3

Output:
1

**Example 2: **

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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.

用DFS/BFS遍历一遍二叉树，要保证左子树比右子树先遍历到，这样没出现一个更深的节点就一定是该深度的最左边的节点。

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/**
 * 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) {
        return BFS(root);
    }
    
    int BFS(TreeNode *root) {
        queue<TreeNode*> nodeQueue;
        queue<int>       depthQueue; // Store the corresponding depth
        nodeQueue.push(root);
        depthQueue.push(0);
        
        int maxDepth = -1;
        int leftBottomVal;
        while (!nodeQueue.empty()) {
            TreeNode *node = nodeQueue.front();
            nodeQueue.pop();
            int depth = depthQueue.front();
            depthQueue.pop();
            if (maxDepth < depth) {
                maxDepth = depth;
                leftBottomVal = node->val;
            }
            if (node->left) {
                nodeQueue.push(node->left);
                depthQueue.push(depth + 1);
            }
            if (node->right) {
                nodeQueue.push(node->right);
                depthQueue.push(depth + 1);
            }
        }
        return leftBottomVal;
    }
};