力扣算法 Java 刷题笔记【二叉树篇】hot100(一)翻转二叉树 填充每个节点的下一个右侧节点指针(中等) 二叉树展开为链表(中等)3
1. 翻转二叉树(简单)
地址: https://leetcode-cn.com/problems/invert-binary-tree/
2021/11/29
AC
写递归算法的关键是要明确函数的「定义」是什么,然后相信这个定义,利用这个定义推导最终结果,绝不要跳入递归的细节
// 将整棵树的节点翻转
TreeNode invertTree(TreeNode root) {
if (root == null) {
return null;
}
TreeNode tmp = root.left;
root.left = root.right;
root.right = tmp;
invertTree(root.left);
invertTree(root.right);
return root;
}
2. 填充每个节点的下一个右侧节点指针(中等)
地址: https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node/
2021/11/29
做题反思: root 为空的条件别忘了,NPE
class Solution {
public Node connect(Node root) {
// root 为空的条件别忘了,NPE
if (root == null) {
return null;
}
connectTwo(root.left, root.right);
return root;
}
void connectTwo(Node a, Node b) {
if (a == null || b == null) {
return;
}
a.next = b;
connectTwo(a.left, a.right);
connectTwo(b.left, b.right);
connectTwo(a.right, b.left);
return;
}
}
3. 二叉树展开为链表(中等)
地址: https://leetcode-cn.com/problems/flatten-binary-tree-to-linked-list/
2021/11/29
做题反思:再做一遍 注意后序遍历代码的逻辑
class Solution {
public void flatten(TreeNode root) {
if (root == null) {
return;
}
flatten(root.left);
flatten(root.right);
TreeNode left = root. left;
TreeNode right = root.right;
root.left = null;
root.right = left;
while (root.right != null) {
root = root.right;
}
root.right = right;
return;
}
}