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.

这个题的思路其实跟[LeetCode] 199. Binary Tree Right Side View_ Medium tag: BFS, Amazon里面我提到的left side view一样的思路, 只是返回的时候返回最后一个元素即可.

1. Constraints

1) root cannot be None, 所以edge case就是 1

2, Ideas

BFS:       T: O(n),   S: O(n)    n is the number of the nodes of the tree

3. Code

 class Solution:
def LeftViewMost(self, root):
ans, queue = [], collections.deque([(root, 0)])
while queue:
node, heig = queue.popleft()
if heig == len(ans):
ans.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return ans[-1]

2)

class Solution(object):
def findBottomLeftValue(self, root):
"""
:type root: TreeNode
:rtype: int
"""
ans, queue = (root.val, 0), collections.deque([(root, 0)])
while queue:
node, height = queue.popleft()
if height > ans[1]:
ans = (node.val, height)
if node.left:
queue.append((node.left, height + 1))
if node.right:
queue.append((node.right, height + 1))
return ans[0]

4. Test case

1) root is 1

2)

    2
/ \
1 3 3)
        1
/ \
2 3
/ / \
4 5 6
/
7

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