[Leetcode] Find Leaves of Binary Tree

Find Leaves of Binary Tree

Given a binary tree, collect a tree's nodes as if you were doing this: Collect and remove all leaves, repeat until the tree is empty.

Example:
Given binary tree

      1
     / \
    2   3
   / \     
  4   5    

Returns [4, 5, 3], [2], [1].

Explanation:

1. Removing the leaves [4, 5, 3] would result in this tree:

    1
   / 

   2

2. Now removing the leaf [2] would result in this tree:

        1          

3. Now removing the leaf [1] would result in the empty tree:

        []         

   Returns [4, 5, 3], [2], [1].

递归

复杂度

O(N) 时间 O(N) 空间

思路

这道题换句话说就是找每个node的index，这个index就是最后结果中这个节点所在的list的index，比如4,5,3的index是0， 2的index是1，1的index是2.
怎么找呢？二分，看左，看右。
确定一个点的index，得知道他的左孩子index是多少，右孩子的index是多少，当前这个点的index是他左右孩子的index最大值+1，这可以很容易地观察得到。比如对于1来说，左孩子2的index是1，右孩子3的index是0，那1的index肯定是max(1, 0) + 1，即2.

注意

if (list.size() == cur) {
    list.add(new ArrayList<Integer>());
}
这段代码为什么可以这么写？
因为通过跟踪递归可以发现，他的index都是连续的，没有跳跃写的情况，一般是这样：先0，再1，再2，再0，再1，再2，再3；不会出现先0，再2，再1，再4这样的情况
    

代码

public class Solution {
    public List<List<Integer>> findLeaves(TreeNode root) {
        List<List<Integer>> list = new ArrayList<>();
        helper(list, root);
        return list;
    }

//calculate the index of this root passed in and put it in that index, at last return where this root was put
    private int helper(List<List<Integer>> list, TreeNode root) {
        if (root == null)
            return -1;
        int left = helper(list, root.left);
        int right = helper(list, root.right);
        int cur = Math.max(left, right) + 1;
        if (list.size() == cur)
            list.add(new ArrayList<Integer>());
        list.get(cur).add(root.val);
        return cur;
    }
}