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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;
}
}
