LeetCode 814. Binary Tree Pruning

原题链接在这里：https://leetcode.com/problems/binary-tree-pruning/

题目：

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.

题解：

DFS from bottom.

If root == null, return null.

Check left and right.

If left != null or right != null or root.val == 1, this means its subtree containing itself has at least 1, return root.

Otherwise, return null.

Time Complexity: O(n).

Space: O(h).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode pruneTree(TreeNode root) {
        if(root == null){
            return root;
        }
        
        root.left = pruneTree(root.left);
        root.right = pruneTree(root.right);
        if(root.val == 1 || root.left != null || root.right != null){
            return root;
        }
        
        return null;
    }
}