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
200 nodes
. - The value of each node will only be
0
or1
.
二叉树剪枝。
给定二叉树根结点 root ,此外树的每个结点的值要么是 0,要么是 1。
返回移除了所有不包含 1 的子树的原二叉树。
( 节点 X 的子树为 X 本身,以及所有 X 的后代。)
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/binary-tree-pruning
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思路是后序遍历。对于每一个节点值 node.val,如果他没有左右孩子或者他的左右孩子的节点值都是 0 的话,就把他剪除(往上返回 null)。可以和1325题放在一起做,两者几乎是同一道题。
时间O(n)
空间O(n)
Java实现
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode() {} 8 * TreeNode(int val) { this.val = val; } 9 * TreeNode(int val, TreeNode left, TreeNode right) { 10 * this.val = val; 11 * this.left = left; 12 * this.right = right; 13 * } 14 * } 15 */ 16 class Solution { 17 public TreeNode pruneTree(TreeNode root) { 18 // corner case 19 if (root == null) { 20 return null; 21 } 22 root.left = pruneTree(root.left); 23 root.right = pruneTree(root.right); 24 if (root.left == null && root.right == null && root.val == 0) { 25 return null; 26 } 27 return root; 28 } 29 }
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