# Find Leaves of Binary Tree

<https://shuati.gitbook.io/crack-lintcode/~/edit/drafts/-LML4yotQUd1ltlwtPxk/binary-tree/find-leaves-of-binary-tree-07-25>

time o(n),space o(n)

mimic the leave level as index,

when we recursively call dfs to the leave, we get the return is 0, if now res.size() <= index, which means we neet to create a new list, otherwise , the corresponding index list is already in the res

return index+1 to the upper level


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