计算给定二叉树的所有左叶子之和。
示例：
3
/ \
9 20
/ \
15 7
在这个二叉树中，有两个左叶子，分别是 9 和 15，所以返回 24
# Definition for a binary tree node.
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
# 递归
def sumOfLeftLeaves(self, root: TreeNode) -> int:
if root is None:return 0
if root.left and root.left.left is None and root.left.right is None:
return root.left.val + self.sumOfLeftLeaves(root.right)
return self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)
# 非递归
def sumOfLeftLeaves2(self, root: 'TreeNode') -> 'int':
if not root:
return 0
queue = [root]
sum_left = 0
while queue:
node = queue.pop(0)
lnode = node.left
if lnode and not lnode.left and not lnode.right:
sum_left += lnode.val
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return sum_left
def sumOfLeftLeaves(self, root: 'TreeNode') -> 'int':
if root is None: return 0
s = 0
if root.left:
if root.left.left is None and root.left.right is None:
s += root.left.val
else:
s += self.sumOfLeftLeaves(root.left)
if root.right:
s += self.sumOfLeftLeaves(root.right)
return s
if __name__ == '__main__':
s = Solution()
root = TreeNode(6)
root.left = TreeNode(2)
root.right = TreeNode(8)
root.left.left = TreeNode(0)
root.left.right = TreeNode(4)
root.left.right.left = TreeNode(3)
root.left.right.right = TreeNode(5)
root.right.left = TreeNode(7)
root.right.right = TreeNode(9)
'''
示例:
给定如下二叉树
6
/ \
2 8
/ \ / \
0 4 7 9
/ \
3 5
'''
print(s.sumOfLeftLeaves(root))