Populating Next Right Pointers in Each Node
Populating Next Right Pointers in Each Node
题目
Given a binary tree
struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *next;
}
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
Initially, all next pointers are set to NULL.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
1
/ \
2 3
/ \ / \
4 5 6 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ / \
4->5->6->7 -> NULL
思路
按图分析，若root节点存在左节点，则左结点的next指向右节点。若root的next不为NULL，则root节点的右节点的next指向root.next节点的左结点。递归下去。
解题
c++版1
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24/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
* int val;
* TreeLinkNode *left, *right, *next;
* TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
if(root == NULL || root -> left == NULL) {
return;
}
root->left->next = root->right;
if(root->next != NULL) {
root->right->next = root->next->left;
}
connect(root->left);
connect(root->right);
return;
}
};
python版1
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19# Definition for binary tree with next pointer.
# class TreeLinkNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# self.next = None
class Solution:
# @param root, a tree link node
# @return nothing
def connect(self, root):
if not root or not root.left:
return
root.left.next=root.right
if root.next:
root.right.next=root.next.left
self.connect(root.left)
self.connect(root.right)
