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
or1
.
解答
解法1:
- 设置递归的反馈的形式,反馈下层是否有可用的(1)存在。
- 参数的设置也很重要,传递哪些信息给下层递归。
- 【小技巧:结构补充】开头添加father是为了后面程序处理的一致性。
public class BinaryTreePruning {
public TreeNode pruneTree(TreeNode root) {
TreeNode father = new TreeNode(1);
father.left = root;
pruneAndJudgeTree(father,true,father.left);
if(father.left == null){
return null;
}else{
return father.left;
}
}
public boolean pruneAndJudgeTree(TreeNode parent,boolean isLeft,TreeNode treeNode){
if(treeNode == null){
return false;
}
boolean isAvailLeft = pruneAndJudgeTree(treeNode,true,treeNode.left);
boolean isAvailRight = pruneAndJudgeTree(treeNode,false,treeNode.right);
if(isAvailLeft || isAvailRight){
return true;
}else{
if(treeNode.val == 1){
return true;
}else{
if(isLeft){
parent.left = null;
}else{
parent.right = null;
}
return false;
}
}
}
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
}