二叉树展开为链表
二叉树展开为链表
LeetCode入口👉👉👉No.114
给定一个二叉树,原地将它展开为一个单链表。
例如,给定二叉树
1
/ \
2 5
/ \ \
3 4 6
将其展开为:
1
\
2
\
3
\
4
\
5
\
6
思路
- 将左子树插入到右子树的地方
- 将原来的右子树接到左子树的最右边节点
- 考虑新的右子树的根节点,一直重复上边的过程,直到新的右子树为 null
1
/ \
2 5
/ \ \
3 4 6
//将 1 的左子树插入到右子树的地方
1
\
2 5
/ \ \
3 4 6
//将原来的右子树接到左子树的最右边节点
1
\
2
/ \
3 4
\
5
\
6
//将 2 的左子树插入到右子树的地方
1
\
2
\
3 4
\
5
\
6
//将原来的右子树接到左子树的最右边节点
1
\
2
\
3
\
4
\
5
\
6
python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def flatten(self, root: TreeNode) -> None:
"""
Do not return anything, modify root in-place instead.
"""
if not root:
return None
while root:
#1.左子树空,考虑下个节点
if not root.left:
root = root.right
#3.右子树接到原左子树的最右节点
else:
pre = root.left
while pre.right:
pre = pre.right
pre.right = root.right
#2.左子树插入到右子树的地方
root.right = root.left
root.left = None
root = root.right
java
class Solution {
public void flatten(TreeNode root) {
//将左子树接到右子树 root.right = root.left
//为了做到这一点,首先要把右子树移走,移到哪?移到左子树的右子树
//pre = root.left; pre.right = root.right;
while(root != null){
//左子树为空,直接考虑下个节点
if(root.left == null){
root = root.right;
}else{
TreeNode pre = root.left;
while(pre.right != null){
pre = pre.right;
}
pre.right = root.right;
root.right = root.left;
root.left = null;
root = root.right;
}
}
}
}
使用前序遍历方法
class Solution {
List<TreeNode> list;
public void flatten(TreeNode root) {
list = new ArrayList<TreeNode>();
preorder(root);
int size = list.size();
for (int i = 1; i < size; i++) {
TreeNode prev = list.get(i - 1), curr = list.get(i);
prev.left = null;
prev.right = curr;
}
}
void preorder(TreeNode root) {
if (root != null) {
list.add(root);
preorder(root.left);
preorder(root.right);
}
}
}