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Find Leaves of Binary Tree in C++
Suppose we have a binary tree. We will collect and remove all leaves and repeat until the tree is empty.
So, if the input is like
then the output will be [[4,5,3],[2],[1]]
To solve this, we will follow these steps −
Define one map sz
Define one 2D array ret
Define a function dfs(), this will take node,
-
if node is null, then −
sz[val of node] := 1 + maximum of dfs(left of node) and dfs(right of node)
-
if size of ret < sz[val of node], then −
Define an array temp
insert temp at the end of ret
insert val of node at the end of ret[sz[val of node] - 1]
return sz[val of node]
From the main method, do the following −
dfs(root)
return ret
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h> using namespace std; void print_vector(vector<vector<auto< > v){ cout << "["; for(int i = 0; i<v.size(); i++){ cout << "["; for(int j = 0; j <v[i].size(); j++){ cout << v[i][j] << ", "; } cout << "],"; } cout << "]"<<endl; } class TreeNode{ public: int val; TreeNode *left, *right; TreeNode(int data){ val = data; left = NULL; right = NULL; } }; void insert(TreeNode **root, int val){ queue<TreeNode*> q; q.push(*root); while(q.size()){ TreeNode *temp = q.front(); q.pop(); if(!temp->left){ if(val != NULL) temp->left = new TreeNode(val); else temp->left = new TreeNode(0); return; }else{ q.push(temp->left); } if(!temp->right){ if(val != NULL) temp->right = new TreeNode(val); else temp->right = new TreeNode(0); return; }else{ q.push(temp->right); } } } TreeNode *make_tree(vector<int< v){ TreeNode *root = new TreeNode(v[0]); for(int i = 1; i<v.size(); i++){ insert(&root, v[i]); } return root; } class Solution { public: unordered_map <int, int> sz; vector < vector <int< > ret; int dfs(TreeNode* node){ if(!node) return 0; sz[node->val] = 1 + max(dfs(node->left), dfs(node->right)); if(ret.size() < sz[node->val]){ vector <int< temp; ret.push_back(temp); } ret[sz[node->val] - 1].push_back(node->val); return sz[node->val]; } vector<vector<int<> findLeaves(TreeNode* root) { dfs(root); return ret; } }; main(){ Solution ob; vector<int< v = {1,2,3,4,5}; TreeNode *root = make_tree(v); print_vector(ob.findLeaves(root)); }
Input
{1,2,3,4,5}
Output
[[3, 5, 4, ],[2, ],[1, ],]
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