- Binary Tree PruningWe 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; }}