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Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

5     / \    4   8   /   / \  11  13  4 /  \    / \7    2  5   1

Return:

[   [5,4,11,2],   [5,8,4,5]]

和112. Path Sum差别不大，多保存一下路径即可。

stack中保存的是当前节点，当前path所有节点的val（这是一个list），和当前path的sum。

class Solution:    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:        if root is None: return root        stack=[(root, [root.val] ,root.val)]        final=[]        while stack:            root, lis, listsum=stack.pop(0)            if root.left is None and root.right is None:                if listsum==sum: final.append(lis)                else: continue            if root.left:                stack.append((root.left, lis+[root.left.val], listsum+root.left.val))            if root.right:                stack.append((root.right, lis+[root.right.val], listsum+root.right.val))        return final

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