[LeetCode] 366. Find Leaves of Binary Tree 找二叉树的叶节点

Given a binary tree, find all leaves and then remove those leaves. Then repeat the previous steps until the tree is empty.

Example:
Given binary tree

          1
         / \
        2   3
       / \
      4   5

Returns [4, 5, 3], [2], [1].

Explanation:

1. Remove the leaves [4, 5, 3] from the tree

          1
         /
        2

2. Remove the leaf [2] from the tree

          1

3. Remove the leaf [1] from the tree

          []

Returns [4, 5, 3], [2], [1].

Credits:
Special thanks to @elmirap for adding this problem and creating all test cases.

给一个二叉树，找出它的叶节点然后删除，重复此步骤，直到二叉树为空。

The key to solve this problem is converting the problem to be finding the index of the element in the result list. Then this is a typical DFS problem on trees.

Java:

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

// traverse the tree bottom-up recursively
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 curr = Math.max(left, right)+1;

    // the first time this code is reached is when curr==0,
    //since the tree is bottom-up processed.
    if(list.size()<=curr){
        list.add(new ArrayList<Integer>());
    }

    list.get(curr).add(root.val);

    return curr;
}　

Python:

class Solution(object):
    def findLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        def findLeavesHelper(node, result):
            if not node:
                return -1
            level = 1 + max(findLeavesHelper(node.left, result), \
                            findLeavesHelper(node.right, result))
            if len(result) < level + 1:
                result.append([])
            result[level].append(node.val)
            return level

        result = []
        findLeavesHelper(root, result)
        return result

C++:

// Time:  O(n)
// Space: O(h)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> findLeaves(TreeNode* root) {
        vector<vector<int>> result;
        findLeavesHelper(root, &result);
        return result;
    }

private:
    int findLeavesHelper(TreeNode *node, vector<vector<int>> *result) {
        if (node == nullptr) {
            return -1;
        }
        const int level = 1 + max(findLeavesHelper(node->left, result),
                                  findLeavesHelper(node->right, result));
        if (result->size() < level + 1){
            result->emplace_back();
        }
        (*result)[level].emplace_back(node->val);
        return level;
    }
};

　　　

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[LeetCode] 104. Maximum Depth of Binary Tree 二叉树的最大深度

[LeetCode] 310. Minimum Height Trees 最小高度树

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