366. Find Leaves of Binary Tree

Given the root of a binary tree, collect a tree’s nodes as if you were doing this:

• Collect all the leaf nodes.
• Remove all the leaf nodes.
• Repeat until the tree is empty.

• new solution update 2022.05.13

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
    vector<vector<int>> res;
    int helper(TreeNode* node) {
        if(!node) return -1;
        int lvl = max(helper(node->left) + 1, helper(node->right) + 1);
        if(res.size() == lvl)
            res.emplace_back();
        res[lvl].push_back(node->val);
        return lvl;
    }
public:
    vector<vector<int>> findLeaves(TreeNode* root) {
        helper(root);
        return res;
    }
};

• dfs solution
• Time : O(n)
• Space : O(1)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
    vector<vector<int>> res;
    bool isLeaf(TreeNode* node) {
        return node->left == nullptr and node->right == nullptr;
    }
    int travel(TreeNode* node) {
        if(!node) return -1;
        
        if(isLeaf(node)) {
            if(res.empty())
                res.push_back(vector<int>());
            res[0].push_back(node->val);
            return 0;
        }
        
        int level = max(travel(node->left), travel(node->right));
        level++;
        while(res.size() <= level) {
            res.push_back(vector<int>());
        }
        res[level].push_back(node->val);
        return level;
    }
public:
    vector<vector<int>> findLeaves(TreeNode* root) {
        travel(root);
        return res;
    }
};

• topology sort solution
• Time : O(n)
• Space : O(n)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
    vector<TreeNode*> leaf;
    unordered_map<TreeNode*, TreeNode*> parent;
    bool isLeaf(TreeNode* node) {
        return !node->left and !node->right;
    }
    
    void search(TreeNode* node) {
        if(isLeaf(node)) {
            leaf.push_back(node);
        }
        if(node->left) {
            parent[node->left] = node;
            search(node->left);
        }
        if(node->right) {
            parent[node->right] = node;
            search(node->right);
        }
    }
    
    TreeNode* getUnLinkParent(TreeNode* node) {
        TreeNode* p = parent[node];
        if(p == NULL) return NULL;
        if(p->left == node) p->left = NULL;
        else p->right = NULL;
        return p;
    }
public:
    vector<vector<int>> findLeaves(TreeNode* root) {
        search(root);
        vector<vector<int>> res;
        while(!leaf.empty()) {
            vector<int> ans;
            vector<TreeNode*> nxt;
            for(auto& node : leaf) {
                ans.push_back(node->val);
                auto parent = getUnLinkParent(node);
                if(parent && isLeaf(parent)) {
                    nxt.push_back(parent);
                }
            }
            nxt.swap(leaf);
            res.push_back(ans);
        }
        return res;
    }
};

• dfs solution
• Time : O(nh)
• Space : O(1)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
    vector<vector<int>> res;
    bool isLeafNode(TreeNode* node) {
        if(!node->left and !node->right){
            res.back().push_back(node->val);
            return true;
        }

        if(node->left and isLeafNode(node->left)) {
            node->left = NULL;
        }
        if(node->right and isLeafNode(node->right)) {
            node->right = NULL;
        }
        return false;
    }
public:
    vector<vector<int>> findLeaves(TreeNode* root) {
        res.push_back(vector<int>());
        while(!isLeafNode(root)) {
            res.push_back(vector<int>());
        }
        return res;
    }
};