标签：优先 遍历 lang sel 广度 otto last print bsp
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.
求二叉树左下方节点的值，使用广度优先遍历，每遍历一层刷新result值
class Solution(object):def findBottomLeftValue(self, root):""":type root: TreeNode:rtype: int"""left = 0if(not root):return leftresult = []queue = [root]while(queue):count = len(queue)for i in range(0, count):node = queue.pop(0)if i == 0:left = node.valif(node.left):queue.append(node.left)if(node.right):queue.append(node.right)return left
513. Find Bottom Left Tree Value 二叉树左下节点的值
标签：优先 遍历 lang sel 广度 otto last print bsp
原文地址：http://www.cnblogs.com/xiejunzhao/p/7355276.html
