算法学习day15

文章目录

- - 102. 二叉树层序遍历
  - - 思路
    - 递归
  - 226 翻转二叉树
  - - 递归
    - 迭代
  - 101 对称二叉树
  - - 递归
  - 总结

102. 二叉树层序遍历

给你二叉树的根节点 root ，返回其节点值的 层序遍历 。（即逐层地，从左到右访问所有节点）

示例 1：
输入：root = [3,9,20,null,null,15,7]
输出：[[3],[9,20],[15,7]]
示例 2：输入：root = [1]
输出：[[1]]
示例 3：输入：root = []
输出：[]

提示：

树中节点数目在范围 [0, 2000] 内
-1000 <= Node.val <= 1000

思路

广度优先，借助队列实现，.
队列的长度也就是每层的元素个数，也就是出队列的个数，即每层的元素值

/*** 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) {List<List<Integer>> res = new ArrayList<>();if (root == null) {return res;}Queue<TreeNode> queue = new LinkedList<>();queue.offer(root);while (!queue.isEmpty()) {int size = queue.size();List<Integer> list = new ArrayList<>();for (int i = 0; i < size; i++) {TreeNode node = queue.poll();list.add(node.val);if (node.left!= null) {queue.offer(node.left);}if (node.right!= null) {queue.offer(node.right);}}res.add(list);}return res;}
}

递归

递归的本质即是“栈”
深度优先遍历

class Solution {List<List<Integer>> res = new ArrayList<>();public List<List<Integer>> levelOrder(TreeNode root) {level(root, 1);return res;}public void level(TreeNode root, int deepth){if(root == null)    return ;//返回列表中的元素个数==层级数if(res.size() < deepth){List<Integer> item = new ArrayList<Integer>();res.add(item);}res.get(deepth-1).add(root.val);//左右节点位于同一层级deepth+1level(root.left, deepth+1);level(root.right, deepth+1);}
}

226 翻转二叉树

给你一棵二叉树的根节点 root ，翻转这棵二叉树，并返回其根节点。

输入：root = [4,2,7,1,3,6,9]
输出：[4,7,2,9,6,3,1]
示例 2：
输入：root = [2,1,3]
输出：[2,3,1]
示例 3：输入：root = []
输出：[]
提示：树中节点数目范围在 [0, 100] 内
-100 <= Node.val <= 100

递归

观察翻转后的结果，实质上时每个节点左右子树的翻转，每个都翻转可联想到递归实现

/*** 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) {//终止条件 节点为nullif(root == null)    return root;//递归内容，交换左右节点TreeNode temp = root.left;root.left = root.right;root.right = temp;//递归入参，每个节点invertTree(root.left);invertTree(root.right);return root;}
}

迭代

迭代遍历中先序遍历：根左右，入栈跟右左，交换左右节点顺序即可

class Solution {public TreeNode invertTree(TreeNode root) {if(root == null) {return null;}   Stack<TreeNode> stack = new Stack<TreeNode>();stack.push(root);while(!stack.isEmpty()) {TreeNode node = stack.pop();if(node.right!= null) {stack.push(node.right);}if(node.left!= null) {stack.push(node.left);}TreeNode temp = node.left;node.left = node.right;node.right = temp;}return root;}
}

101 对称二叉树

给你一个二叉树的根节点 root ，检查它是否轴对称。

输入：root = [1,2,2,3,4,4,3] 输出：true输入：root = [1,2,2,null,3,null,3]
输出：false
提示：树中节点数目在范围 [1, 1000] 内
-100 <= Node.val <= 100

递归

对称，左右子树的内测和外侧要分别相同，左子树的左侧右子树的右侧；左子树的右侧右子树的左侧
递归入参：内外侧子树
终止条件：有一个子树为空，或者值不相等

/*** 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 isSymmetric(TreeNode root) {return checkLeftEquelsRight(root.left, root.right);}public boolean checkLeftEquelsRight(TreeNode left, TreeNode right) {if(left ==null && right ==null){return true;}if(right!=null && left == null){return false;}if(left!=null && right==null){return false;}if(left.val !=  right.val) {return false;}return checkLeftEquelsRight(left.left,right.right) && checkLeftEquelsRight(left.right,right.left);}}