814. Binary Tree Pruning**
https://leetcode.com/problems/binary-tree-pruning/
题目描述
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.
C++ 实现 1
由于前面做了 1325. Delete Leaves With a Given Value** 这道题, 写这题时很快就搞定了. 思路 1, 对左右子树进行裁剪, 然后判断根节点自身是否符合被裁剪的条件. 显然, 这是后序遍历.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* pruneTree(TreeNode* root) {
if (!root)
return nullptr;
root->left = pruneTree(root->left);
root->right = pruneTree(root->right);
return (!root->left && !root->right && root->val == 0) ? nullptr : root;
}
};
C++ 实现 2
参考 1325. Delete Leaves With a Given Value** 中的 C++ 实现 2
可以了解, 如果使用指针的引用, 可以在遍历过程中对树进行修改.
依然是后序遍历的思路, 如果遇到 root 为叶子节点并且 roo->val == 0
, 就可以将 root 设置为 nullptr
.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
void pruning(TreeNode *&root) {
if (!root) return;
pruning(root->left);
pruning(root->right);
if (!root->left && !root->right && root->val == 0) root = nullptr;
}
public:
TreeNode* pruneTree(TreeNode* root) {
pruning(root);
return root;
}
};