Java
/**
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.
*/
public class Binary_Tree_Pruning {
/**
* 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);
// leaf node
if (root.left == null && root.right == null) {
if (root.val == 0) {
return null;
} else {
return root;
}
}
return root;
}
}
}Java
/**
* 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;
Map<TreeNode, TreeNode> childParentMap = new HashMap<TreeNode, TreeNode>();
List<TreeNode> nodesList = new ArrayList<TreeNode>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
nodesList.add(node);
TreeNode left = node.left, right = node.right;
if (left != null) {
childParentMap.put(left, node);
queue.offer(left);
}
if (right != null) {
childParentMap.put(right, node);
queue.offer(right);
}
}
for (int i = nodesList.size() - 1; i >= 0; i--) {
TreeNode node = nodesList.get(i);
if (node.val == 0) {
if (node.left == null && node.right == null) {
TreeNode parent = childParentMap.get(node);
if (parent != null) {
if (node == parent.left)
parent.left = null;
else
parent.right = null;
} else {
root = null;
break;
}
}
}
}
return root;
}
}