LC-Sum of Left Leaves
2018-02-12 23:39
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# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def sumOfLeftLeaves(self, root): """ :type root: TreeNode :rtype: int """ if not root: return 0 if root.left and not root.left.left and not root.left.right: return root.left.val + self.sumOfLeftLeaves(root.right) return self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)
0,参考后的代码
1,题目要求:
也就是计算一棵树中所有的左叶子节点的值的和。
2,依旧是以一种递归的算法,需要注意的是对于节点的条件判断:当节点为空时,返回值为0。对于一个节点,如果其左孩子不为空但是左孩子没有孩子,说明该节点的左孩子为左叶子节点,也即符合题目条件的节点。按照要求,取该节点左孩子的值并加上该节点右孩子的递归算法(找右子树中的左叶节点)。
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