34. 二叉树中和为某一值的路径

Posted 2021-03-01 ustc-zcc

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
        res=[]
        def traceback(node,trace,sum):
            if not node:
                return 
            if node.val==sum and (not node.left) and (not node.right):
                res.append(trace+[node.val])
                return
            if node.left:
                traceback(node.left,trace+[node.val],sum-node.val)
            if node.right:
                traceback(node.right,trace+[node.val],sum-node.val)
        traceback(root,[],sum)
        return res

 

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
        res, path = [], []

        def dfs(node, sum):
            #递归出口：解决子问题
            if not node: return #如果没有节点(node = None)，直接返回，不向下执行
            else:               #有节点
                path.append(node.val) #将节点值添加到path
                sum -= node.val 
            # 如果节点为叶子节点，并且 sum == 0
            if not node.left and not node.right and not sum: 
                res.append(path[:]) 

            dfs(node.left, sum) #递归处理左边
            dfs(node.right, sum) #递归处理右边
            path.pop() #处理完一个节点后，恢复初始状态，为node.left,  node.right操作

        dfs(root, sum)
        return res