题目描述：
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.
解题思路：
遍历整棵树，记录每个节点的值和深度，得到最大深度值，第一个深度值等于最大深度值的节点为所要求的节点。
代码：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int findBottomLeftValue(TreeNode* root) { 13 bfs(root, 0); 14 int max_height = -1; 15 for (auto h : height) { 16 if (h > max_height) 17 max_height = h; 18 } 19 int i = 0; 20 for (; i < height.size(); ++i) { 21 if (height[i] == max_height) { 22 break; 23 } 24 } 25 return node[i]; 26 } 27 vector<int> node; 28 vector<int> height; 29 void bfs(TreeNode* root, int h) { 30 if (root == NULL) 31 return; 32 if (root->left != NULL) 33 bfs(root->left, h+1); 34 if (root->right != NULL) 35 bfs(root->right, h+1); 36 node.push_back(root->val); 37 height.push_back(h); 38 } 39 };