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题目:
We are given the head node
root
of a binary tree, where additionally every node’s value is either a0
or a1
.
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 most100
nodes.
The value of each node will only be0
or1
.
解释:
把不包括1的子树移除。
递归写法:
python代码:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def pruneTree(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if not root:
return None
if root.left:
root.left=self.pruneTree(root.left)
if root.right:
root.right=self.pruneTree(root.right)
if not root.left and not root.right and not root.val:
return None
return root
c++代码:
/**
* 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 NULL;
if (root->left)
root->left=pruneTree(root->left);
if(root->right)
root->right=pruneTree(root->right);
if (!root->left &&!root->right && !root->val)
return NULL;
return root;
}
};
总结:
可以用函数本身递归的,就不需要在写一个helper()
函数了。