Python实现"左叶子之和"的两种方法

给定一颗二叉树，返回它所有左叶子节点之和

Example:

    3
   / \
  9  20
    /  \
   15   7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

１：递归方法

def sumOfLeftLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        if not root.left and not root.right:   #当前节点不存在左右子树
            return 0
        if not root.left and root.right:       #当前节点只有右子树
            return self.sumOfLeftLeaves(root.right)
        if root.left and not root.right:     #当前节点只有左子树
            if not root.left.left and not root.left.right:   #当前节点的左节点不存在左右子树
                return root.left.val
            else:
                return self.sumOfLeftLeaves(root.left)
        sum = 0
        if root.left and root.right:        #当前节点既有左子树又有右子树
            if not root.left.left and not root.left.right:         #当前节点的左子树没有左右节点
                sum += root.left.val
            else:
                sum += self.sumOfLeftLeaves(root.left)
            sum += self.sumOfLeftLeaves(root.right)
        return sum

简化写法（参考他人）

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

２：迭代方法

def sumOfLeftLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        sum = 0
        node = [root]      #存放节点
        while node:
            cur = node.pop(0)         #当前节点
            if not cur.left and not cur.right:   #当前节点无左右子树
                continue
            elif cur.left and not cur.right:        #当前节点有左子树无右子树
                if not cur.left.left and not cur.left.right:     #当前节点的左子树无左右子树，满足左叶子节点条件
                    sum += cur.left.val
                else:
                    node.append(cur.left)
            elif not cur.left and cur.right:          #当前节点无左子树有右子树
                node.append(cur.right)
            else:        #当前节点有左右子树
                if not cur.left.left and not cur.left.right:       #当前节点的左子树无左右子树，满足左叶子节点条件
                    sum += cur.left.val
                else:
                    node.append(cur.left)
                node.append(cur.right)
        return sum

简化写法（参考他人）

def sumOfLeftLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        sum = 0
        node = [root]      #存放节点
        while node:
            cur = node.pop(0)         #当前节点
            if cur.left:
                if not cur.left.left and not cur.left.right:
                    sum += cur.left.val
                else:
                    node.append(cur.left)
            if cur.right:
                node.append(cur.right)
        return sum

算法题来自：https://leetcode-cn.com/problems/sum-of-left-leaves/description/