題目

給定一顆二叉樹，返回它所有左葉子節(jié)點(diǎn)之和

舉例

    3
   / \
  9  20
    /  \
   15   7
return 24

遞歸方法

def sumOfLeftLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        if not root.left and not root.right:   #當(dāng)前節(jié)點(diǎn)不存在左右子樹
            return 0
        if not root.left and root.right:       #當(dāng)前節(jié)點(diǎn)只有右子樹
            return self.sumOfLeftLeaves(root.right)
        if root.left and not root.right:     #當(dāng)前節(jié)點(diǎn)只有左子樹
            if not root.left.left and not root.left.right:   #當(dāng)前節(jié)點(diǎn)的左節(jié)點(diǎn)不存在左右子樹
                return root.left.val
            else:
                return self.sumOfLeftLeaves(root.left)
        sum = 0
        if root.left and root.right:        #當(dāng)前節(jié)點(diǎn)既有左子樹又有右子樹
            if not root.left.left and not root.left.right:         #當(dāng)前節(jié)點(diǎn)的左子樹沒有左右節(jié)點(diǎn)
                sum += root.left.val
            else:
                sum += self.sumOfLeftLeaves(root.left)
            sum += self.sumOfLeftLeaves(root.right)
        return sum

簡(jiǎn)化遞歸寫法（參考他人）

def sumOfLeftLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        sum = 0
        if root.left and not root.left.left and not root.left.right:
            sum += root.left.val
        sum += self.sumOfLeftLeaves(root.left)
        sum += self.sumOfLeftLeaves(root.right)
        return sum