814. Binary Tree Pruning

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.

题目地址

这道题考察的应该是二叉树的遍历，核心的解法在于先将叶子0节点变为null，然后就会发现祖先节点可以用同样的方法递归的解决。

java代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
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==0 && root.left==null && root.right==null){
            root=null;
        }
        return root;
    }
}