102. Binary Tree Level Order Traversal
Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]
Example 2:
Input: root = [1]
Output: [[1]]
Example 3:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 2000]. -1000 <= Node.val <= 1000
BFS 复习。主要就两个要点：
- 用queue推入当前层的子节点，方法是pop一个推一个的子节点，双层循环是必须的
- 用一个size保存当前层的节点数，外层循环每循环处理一层。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
Deque<TreeNode> dq = new LinkedList<TreeNode>();
int size = 0;
List<List<Integer>> ans = new LinkedList<List<Integer>>();
if (root == null) return ans;
dq.offer(root);
while(!dq.isEmpty()) {
size = dq.size();
List<Integer> clist = new LinkedList<Integer>();
while(size-- > 0) {
TreeNode node = dq.pop();
clist.add(node.val);
if(node.left != null) {
dq.offer(node.left);
}
if(node.right != null) {
dq.offer(node.right);
}
}
ans.add(clist);
}
return ans;
}
}
226. Invert Binary Tree
Given the root of a binary tree, invert the tree, and return its root.
Example 1:
Input: root = [4,2,7,1,3,6,9]
Output: [4,7,2,9,6,3,1]
Example 2:
Input: root = [2,1,3]
Output: [2,3,1]
Example 3:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 100]. -100 <= Node.val <= 100
翻转二叉树。这题很简单但要想清楚。前序，后序遍历都可以一次翻完，但中序会重复翻转一些节点。BFS也可以一次翻完，迭代也可以。主要是翻转左右节点的操作中间不能夹推栈操作。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
if(root == null) {
return null;
}
Stack<TreeNode> st = new Stack<TreeNode>();
st.push(root);
while(!st.isEmpty()) {
TreeNode cur = st.pop();
TreeNode tmp = cur.left;
cur.left = cur.right;
cur.right = tmp;
if(cur.left != null) {
st.push(cur.left);
}
if(cur.right != null) {
st.push(cur.right);
}
}
return root;
}
}
101. Symmetric Tree
Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
Example 1:
Input: root = [1,2,2,3,4,4,3]
Output: true
Example 2:
Input: root = [1,2,2,null,3,null,3]
Output: false
Constraints:
- The number of nodes in the tree is in the range
[1, 1000]. -100 <= Node.val <= 100
这题也是要想清楚遍历的调用顺序。既不是前序也不是中后序。本质上算是左右子树的比较，左子树的前序遍历集合要跟右子树的后序遍历集合一致。 直观感受是分别做前序和后序遍历，输出结果再比较，但一次遍历也是可以的。判断条件会比较繁杂，要注意空节点判断。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean rcHelper(TreeNode ln, TreeNode rn) {
if(ln == null || rn == null) {
if(ln == null && rn != null || ln != null && rn == null) {
return false;
}
else {
return true;
}
}
//System.out.println(ln.val + "|" + rn.val);
if(ln.val != rn.val) {
return false;
}
boolean bleft = rcHelper(ln.left, rn.right);
boolean bright = rcHelper(ln.right, rn.left);
return bleft && bright;
}
public boolean isSymmetric(TreeNode root) {
if(root == null) {
return false;
}
return rcHelper(root.left, root.right);
}
}
