Print and remove leaf nodes until root node left

  • 1

    Given a imbalance binary tree, print the leaf nodes then remove those leaf node, print the new leaf nodes until only root node left.

    For example:

             / \
            2   3
           / \     
          4   5    

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

  • 0

    @evanl1208 what is meant by drop? Shall we remove leaves first?

  • 0

    @elmirap yes, I mean remove those leaf nodes.

  • 0

    @evanl1208 thank you

  • 1
    int height(TreeNode * root)
        if(!root) return 0;
        return 1+max(height(root->left),height(root->right));
    int dfs(vector<vector<int>> &res,TreeNode * root)
        if(!root) return 0;
        int n=1+max(dfs(res,root->left),dfs(res,root->right));
        return n;
    vector<vector<int>> func(TreeNode * root)
        int n=height(root);
        vector<vector<int>> res(n);
        return res;
        TreeNode* a1=new TreeNode(3);
        a1->left =new TreeNode(1);
        a1->left->right =new TreeNode(2);
        a1->right =new TreeNode(5);
        a1->right->left =new TreeNode(4);
        a1->right->right =new TreeNode(7);
        a1->right->right->left =new TreeNode(6);
        a1->left->left =new TreeNode(0);
        a1->left->left->left =new TreeNode(-1);

    O(n) time and space, bottom up, hope it is correct!

  • 2

    Very good idea GoGoDong! My idea is the same but I optimize a little as avoiding calculation of getHeight in advance.

    int dropByLevel(TreeNode root,Map<Integer,ArrayList<TreeNode>> map) {
        if (root != null) {
            int l = dropByLevel(root.left, map);
            int r = dropByLevel(root.right, map);
            int h = Math.max(l, r) + 1;
            ArrayList<TreeNode> arr = map.containsKey(h) ? map.get(h)  : new ArrayList<>():;       
            map.put(h, arr);
            return h;
        return 0;
    void test (TreeNode root)  {
    Map<Integer,ArrayList<TreeNode>> map  =  new HashMap<>();     
    dropByLevel(root,  map) ;
    for (Entry<Integer, ArrayList<TreeNode>> entry  :  map.entrySet())  {         

  • 0

    @GoGoDong but you are not removing them? you are only printing.

  • 0

    @evanl1208 Does the order of printing leaves matter?

  • 0

    Recursive approach in Java:

    It is similar to the ones above. Keeps track of the level and deletes the leaf from the parent (set parent.left or parent.right as null and let garbage collector collect the child node). Add the child node's value to the list at index level

    public class PrintAndRemoveLeafNodes {
    	private static List<List<Integer>> leaves;
    	public static List<List<Integer>> printAndRemoveLeafNodes(TreeNode root) {
    		leaves = new ArrayList<List<Integer>>();
    		if (root == null)
    			return leaves;
                    // use a dummy node as parent for root
    		TreeNode dummy = new TreeNode(-1);
    		dummy.left = root;
    		printAndRemoveLeafNodes(dummy, root, true);
    		return leaves;
    	private static int printAndRemoveLeafNodes(TreeNode parent, TreeNode child, boolean left) {
    		int level = 0;
    		if (child.left != null)
    			level = printAndRemoveLeafNodes(child, child.left, true);
    		if (child.right != null)
    			level = Math.max(level, printAndRemoveLeafNodes(child, child.right, false));
                    // know the level at this point 
                    // remove the leaf
    		if (left)
    			parent.left = null;
    			parent.right = null;
    		if (leaves.size() < level + 1)
    			leaves.add(new ArrayList<Integer>());
                    // add leaf to list of that level
    		return level + 1;

  • 0

    would a queue work? a bfs till you get to a leaf node. delete it then at the end just recur the root node. stop when root is nullptr??????

  • 0

    First, let's know the parent of each node, and the number of children they have. We can do this with a simple DFS.

    Now perform a bottom up traversal of leaves using a BFS. A leaf has no children. When we process a node (at this point it is a leaf), its parent now has one less child, and if it is a leaf, enqueue. We will visit the nodes in the correct order, and variable depth will tell us where to put the answer.

    class Node:
        def __init__(self, val):
            self.val = val
            self.left = None
            self.right = None
    def solve(root):
        num_children = collections.Counter()
        parent = {}
        def dfs(node, par = None):
            if node:
                num_children[node] = bool(node.left) + bool(node.right)
                parent[node] = par
                dfs(node.left, node)
                dfs(node.right, node)
        leaves = [(node, 0) for node, num in num_children.iteritems() if num == 0]
        ans = []
        while leaves:
            node, depth = leaves.pop()
            while len(ans) <= depth:
            par = parent[node]
            num_children[par] -= 1
            if num_children[par] == 0:
                leaves.append((par, depth+1))
        return ans

  • 0

    Simple traversal

    Complexity -

    Worst Case - assume tree is slanted either to right or left, each time leaf is popped, a size of tree reduces by 1, so

    T(n) = n + n-1 + n-2 .... + 1 = n(n+1)/2 = O(n^2)

    Best Case - tree is complete balanced binary tree, each time leaves are popped size reduces by half

    T(n) = n + n/2 + n/4 + ..... = O ( n*log(n) )

    class TreeNode(object):
        def __init__(self,val):
            self.val  = val
            self.left = self.right = None
    def popLeaves(node,leaves):
        if node:
            if node.left is None and node.right is None:
                leaves.append(node.val) # node is a leaf, save it in a list
                return None # unlinking it from tree
                # node is internal node
                node.left = popLeaves(node.left, leaves)
                node.right = popLeaves(node.right, leaves)
        return node
    def purge(root):
        result = []
        while root:
            leaves = []
            root  = popLeaves(root,leaves) # pop leaves in current state of a tree
            result.append(leaves[:]) # put all collected leaves in a result
        return result

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