Medium
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.

居然又等到一道不看答案做出来的题。这道题我就用的BFS的level order traversal, 取最后一层的第一个元素返回就好了。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int findBottomLeftValue(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()){
            int size = queue.size();
            List<Integer> level = new ArrayList<>();
            for (int i = 0; i < size; i++){
                TreeNode curt = queue.poll();
                level.add(curt.val);
                if (curt.left != null){
                    queue.offer(curt.left);
                } 
                if (curt.right != null){
                    queue.offer(curt.right);
                }
            }
            res.add(level);
        }
        return res.get(res.size() - 1).get(0);    
    }
}

当然也可以稍微地简化一下,不用保存整个traversal的List,每次只需要记录每一层第一个元素（即最左边）的元素就可以了，用到了queue.peek()这个方法。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int findBottomLeftValue(TreeNode root) {
        int res = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()){
            int size = queue.size();
            res = queue.peek().val;
            for (int i = 0; i < size; i++){
                TreeNode curt = queue.poll();
                if (curt.left != null){
                    queue.offer(curt.left);
                } 
                if (curt.right != null){
                    queue.offer(curt.right);
                }
            }
        }
        return res;    
    }
}