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.

1028_2.png
Example 2:
Input: [1,0,1,0,0,0,1]
Output: [1,null,1,null,1]
1028_2.png
Example 3:
Input: [1,1,0,1,1,0,1,0]
Output: [1,1,0,1,1,null,1]
1028.png
Note:

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

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思路：

该题要求的是截取掉全部含零的子树。我的思路是采用递归的方法查找满足条件的子树。当树结点的值为1的时候返回false。检查树中的点的左右子树。一旦返回的是true就将相应的左右子树设置为null。

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代码：

    public TreeNode pruneTree(TreeNode root) {
        if(root == null){
            return null;
        }
        TreeNode result = root;
        traverse(result, true);
        return result;
    }
    
    private boolean traverse(TreeNode root, boolean flag){
        if(root == null){
            return flag;
        }
        boolean flag1 = traverse(root.left, flag);
        if(flag1){
            root.left = null;
        }
        boolean flag2 = traverse(root.right, flag);
        if(flag2){
            root.right = null;
        }
        if(root.val == 1){
            return false;
        }
        return true && flag1 && flag2;
    }