Question

Formatted question description: https://leetcode.ca/all/366.html

 366	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:

 Input: [1,2,3,4,5]

    1
   / \
  2   3
 / \
4   5

 Output: [[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:

 []

 @tag-tree

Algorithm

Each node is separated from the left child node and the right child node to get two depths,

Since the condition of becoming a leaf node is that the left and right child nodes are empty, we take the larger value of the left and right child nodes plus 1 as the depth value of the current node.

Knowing the depth value, you can add the node value to the correct position in the result.

Code

Java

import java.util.ArrayList;
import java.util.List;

public class Find_Leaves_of_Binary_Tree {

    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; // @note: +1==0, mapping to list index
        }

        int left = helper(list, root.left);
        int right = helper(list, root.right);
        int currentDepthFromBottom = 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() <= currentDepthFromBottom) {
            list.add(new ArrayList<Integer>());
        }

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

        return currentDepthFromBottom;
    }
}

// OJ: https://leetcode.com/problems/find-leaves-of-binary-tree/
// Time: O(N)
// Space: O(H)
class Solution {
private:
    bool dfs(TreeNode *root, vector<int> &v) {
        if (!root) return true;
        if (!root->left && !root->right) {
            v.push_back(root->val);
            return true;
        }
        if (dfs(root->left, v)) root->left = NULL;
        if (dfs(root->right, v)) root->right = NULL;
        return false;
    }
    vector<int> removeLeaves(TreeNode *root) {
        vector<int> v;
        dfs(root, v);
        return v;
    }
public:
    vector<vector<int>> findLeaves(TreeNode* root) {
        if (!root) return {};
        vector<vector<int>> ans;
        while (root->left || root->right) {
            ans.push_back(removeLeaves(root));
        }
        ans.push_back({ root->val });
        return ans;
    }
};

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
import collections


class Solution(object):
  def findLeaves(self, root):
    """
    :type root: TreeNode
    :rtype: List[List[int]]
    """

    def helper(p, res):
      if not p:
        return 0
      left = helper(p.left, res)
      right = helper(p.right, res)
      depth = max(left, right) + 1
      res[depth].append(p.val)
      return depth

    ans = []
    res = collections.defaultdict(list)
    helper(root, res)
    for i in range(1, len(res) + 1):
      ans.append(res[i])
    return ans

366 - Find Leaves of Binary Tree

Question

Algorithm

Code

All Problems

All Solutions