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

Example 1:

Input:    2   / /  1   3Output:1
Example 2: 
Input:        1       / /      2   3     /   / /    4   5   6       /      7Output:7
Note: You may assume the tree (i.e., the given root node) is not NULL.
Subscribe to see which companies asked this question.這應該就是一個很直白的廣搜，每一層每一層的處理，最后返回最后一層的最左邊的元素
# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def findBottomLeftValue(self, root):        s = [root]        t = []        res = root.val        while 1:            t = []            for root in s:                if root.left:                    t += root.left,                if root.right:                    t += root.right,            if t == []:                break            else:                s = t[:]                res = t[0].val        return res        """        :type root: TreeNode        :rtype: int        """
這種方法看起來也很厲害的樣子
def findLeftMostNode(self, root):     queue = [root]     for node in queue:          queue += filter(None(node.right,node.left))     returnnode.val