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.

分析:广搜,层序遍历,保存每层的第一个值,最后的即为所求值。

 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         int result = root->val;
14         queue<TreeNode *> q;
15         q.push(root);
16         TreeNode* temp;
17         while (!q.empty()) {
18             int size = q.size();
19             result = q.front()->val;
20             while (size--) {
21                 temp = q.front();
22                 q.pop();
23                 if (temp->left != NULL) q.push(temp->left);
24                 if (temp->right != NULL) q.push(temp->right);
25             }
26         }
27         return result;
28     }
29 };

 

越努力,越幸运