day17| 110.平衡二叉树；257.二叉树的所有路径；404.左叶子之和-526互联

110. 平衡二叉树

自顶向下递归

1. 获得计算二叉树高度的函数

2. 对于遍历到的节点，首先计算左右子树的高度，看是否平衡

3. 在分别遍历到左右子树，判断左子树和右子树是否平衡

代码如下：

class Solution:
    def isBalanced(self, root: TreeNode) -> bool:
        def height(root: TreeNode) -> int:
            if not root:
                return 0
            return max(height(root.left), height(root.right)) + 1

        if not root:
            return True
        return abs(height(root.left) - height(root.right)) <= 1 and self.isBalanced(root.left) and self.isBalanced(root.right)

257. 二叉树的所有路径

深度优先搜索

class Solution:
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        def construct_paths(root, path):
            if root:
                path += str(root.val)
                if not root.left and not root.right:  # 当前节点是叶子节点
                    paths.append(path)  # 把路径加入到答案中
                else:
                    path += '->'  # 当前节点不是叶子节点，继续递归遍历
                    construct_paths(root.left, path)
                    construct_paths(root.right, path)

        paths = []
        construct_paths(root, '')
        return paths

广度优先搜索

class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        paths = list()
        if not root:
            return paths

        node_queue = collections.deque([root])
        path_queue = collections.deque([str(root.val)])

        while node_queue:
            node = node_queue.popleft()
            path = path_queue.popleft()

            if not node.left and not node.right:
                paths.append(path)
            else:
                if node.left:
                    node_queue.append(node.left)
                    path_queue.append(path + '->' + str(node.left.val))
                
                if node.right:
                    node_queue.append(node.right)
                    path_queue.append(path + '->' + str(node.right.val))
        return paths

404. 左叶子之和

深度优先搜索

class Solution:
    def sumOfLeftLeaves(self, root: TreeNode) -> int:
        isLeafNode = lambda node: not node.left and not node.right

        def dfs(node: TreeNode) -> int:
            ans = 0
            if node.left:
                ans += node.left.val if isLeafNode(node.left) else dfs(node.left)
            if node.right and not isLeafNode(node.right):
                ans += dfs(node.right)
            return ans
        
        return dfs(root) if root else 0

广度优先搜索

class Solution:
    def sumOfLeftLeaves(self, root: TreeNode) -> int:
        if not root:
            return 0
        
        isLeafNode = lambda node: not node.left and not node.right
        q = collections.deque([root])
        ans = 0

        while q:
            node = q.popleft()
            if node.left:
                if isLeafNode(node.left):
                    ans += node.left.val
                else:
                    q.append(node.left)
            if node.right:
                if not isLeafNode(node.right):
                    q.append(node.right)
        
        return ans