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:和637. Average of Levels in Binary Tree
(https://www.jianshu.com/p/814d871c5f6d)的思路差不多,都是层遍历,若要返回最后一层最左侧叶节点,只需相反顺序遍历即可,637题目中的解法是先左子后右子,本题反过来,先右后左,则层遍历最后访问的一定是最后一层最左叶子,直接返回即可.
class Solution {
public:
int findBottomLeftValue(TreeNode* root) {
queue<TreeNode*> q; //节点队列
q.push(root); //根先入队
TreeNode* tmp; //层遍历工作节点
while(!q.empty()) { //每次循环处理一层
tmp = q.front();
q.pop();
//先右后左
if (tmp->right) q.push(tmp->right);
if (tmp->left) q.push(tmp->left);
}
return tmp->val; //循环访问的最后一个结点一定是最底层最左
}
};
