[LeetCode] Find Leaves of Binary Tree 找二叉樹的葉節點

 

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:

          []         

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

 

這道題給了我們一個二叉樹，讓我們返回其每層的葉節點，就像剝洋蔥一樣，將這個二叉樹一層一層剝掉，最后一個剝掉根節點。那么題目中提示說要用DFS來做，思路是這樣的，每一個節點從左子節點和右子節點分開走可以得到兩個深度，由於成為葉節點的條件是左右子節點都為空，所以我們取左右子節點中較大值加1為當前節點的深度值，知道了深度值就可以將節點值加入到結果res中的正確位置了，求深度的方法我們可以參見 Maximum Depth of Binary Tree 中求最大深度的方法，參見代碼如下：

 

解法一：

class Solution {
public:
    vector<vector<int>> findLeaves(TreeNode* root) {
        vector<vector<int>> res;
        helper(root, res);
        return res;
    }
    int helper(TreeNode* root, vector<vector<int>>& res) {
        if (!root) return -1;
        int depth = 1 + max(helper(root->left, res), helper(root->right, res));
        if (depth >= res.size()) res.resize(depth + 1);
        res[depth].push_back(root->val);
        return depth;
    }
};

 

下面這種DFS方法沒有用計算深度的方法，而是使用了一層層剝離的方法，思路是遍歷二叉樹，找到葉節點，將其賦值為NULL，然后加入leaves數組中，這樣一層層剝洋蔥般的就可以得到最終結果了：

 

解法二：

class Solution {
public:
    vector<vector<int>> findLeaves(TreeNode* root) {
        vector<vector<int>> res;
        while (root) {
            vector<int> leaves;
            root = remove(root, leaves);
            res.push_back(leaves);
        }
        return res;
    }
    TreeNode* remove(TreeNode* node, vector<int>& leaves) {
        if (!node) return NULL;
        if (!node->left && !node->right) {
            leaves.push_back(node->val);
            return NULL;
        }
        node->left = remove(node->left, leaves);
        node->right = remove(node->right, leaves);
        return node;
    }
};

 

還有一種不用建立新的遞歸函數的方法，就用本身來做遞歸，我們首先判空，然后對左右子結點分別調用遞歸函數，這樣我們suppose左右子結點的所有葉結點已經按順序存好到了二維數組left和right中，現在要做的就是把兩者合並。但是我們現在並不知道左右子樹誰的深度大，我們希望將長度短的二維數組加入到長的里面，那么就來比較下兩者的長度，把長度存到結果res中，把短的存入到t中，然后遍歷短的，按順序都加入到結果res里，好在這道題沒有強行要求每層的葉結點要按照從左到右的順序存入。當左右子樹的葉結點融合完成了之后，當前結點也要新開一層，直接自己組一層，加入結果res中即可，參見代碼如下：

 

解法三：

class Solution {
public:
    vector<vector<int>> findLeaves(TreeNode* root) {
        if (!root) return {};
        vector<vector<int>> left = findLeaves(root->left), right = findLeaves(root->right);
        vector<vector<int>> res = (left.size() >= right.size()) ? left : right;
        vector<vector<int>> t = (left.size() >= right.size()) ? right : left;
        for (int i = 0; i < t.size(); ++i) {
            res[i].insert(res[i].begin(), t[i].begin(), t[i].end());
        }
        res.push_back({root->val});
        return res;
    }
};

 

類似題目：

Maximum Depth of Binary Tree

Minimum Height Trees

 

參考資料：

https://leetcode.com/problems/find-leaves-of-binary-tree/

https://leetcode.com/problems/find-leaves-of-binary-tree/discuss/83773/1-ms-Easy-understand-Java-Solution

https://leetcode.com/problems/find-leaves-of-binary-tree/discuss/191609/10%2B-line-Java-solution-using-recursion

https://leetcode.com/problems/find-leaves-of-binary-tree/discuss/83778/10-lines-simple-Java-solution-using-recursion-with-explanation

 

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