BinaryTreePruning [source code]

public class BinaryTreePruning {
static
/******************************************************************************/
class Solution {
    public TreeNode pruneTree(TreeNode root) {
        containsOne (root);
        return root;
    }

    boolean containsOne (TreeNode root) {
        if (root == null)
            // tricky base case
            return false;
        boolean left = containsOne (root.left), right = containsOne (root.right);
        if (!left) {
            root.left = null;
        }
        if (!right)
            root.right = null;
        return left || right || root.val == 1;
    }
}
/******************************************************************************/

    public static void main(String[] args) {
        BinaryTreePruning.Solution tester = new BinaryTreePruning.Solution();
    }
}

除了leaf的base case稍微需要思考一下, 本身题目不是很难, 想到mutation recurison和returned recursion混合使用就行了;

UNFINISHED

uwi:

class Solution {  
    public TreeNode pruneTree(TreeNode root) {  
        return dfs(root);  
    }  

    TreeNode dfs(TreeNode cur)  
    {  
        if(cur == null)return null;  
        boolean hasone = false;  
        if(cur.left != null){  
            cur.left = dfs(cur.left);  
            if(cur.left != null)hasone = true;  
        }  
        if(cur.right != null){  
            cur.right = dfs(cur.right);  
            if(cur.right != null)hasone = true;  
        }  
        if(hasone || cur.val == 1){  
            return cur;  
        }else{  
            return null;  
        }  
    }  
}

没仔细看, 不过这个题目本身也不难;

cchao:

class Solution {  
public:  
    TreeNode* pruneTree(TreeNode* root) {  
        if (root == nullptr) return root;  
        root->left = pruneTree(root->left);  
        root->right = pruneTree(root->right);  
        if (root->left == nullptr && root->right == nullptr && root->val == 0)  
            return nullptr;  
        return root;  
    }  
};

Problem Description

We are given the head node root of a binary tree, where additionally every node's value is either a 0 or a 1.

Return the same tree where every subtree (of the given tree) not containing a 1 has been removed.

(Recall that the subtree of a node X is X, plus every node that is a descendant of X.)

Example 1:  
Input: [1,null,0,0,1]  
Output: [1,null,0,null,1]

Explanation:
Only the red nodes satisfy the property "every subtree not containing a 1".
The diagram on the right represents the answer.

Example 2:  
Input: [1,0,1,0,0,0,1]  
Output: [1,null,1,null,1]

Example 3:  
Input: [1,1,0,1,1,0,1,0]  
Output: [1,1,0,1,1,null,1]

Note:

• The binary tree will have at most 100 nodes.
• The value of each node will only be 0 or 1.

Difficulty:Medium
Total Accepted:1.3K
Total Submissions:1.7K
Contributor:awice
Companies
hulu
Related Topics
tree