[代码随想录] 第十三天

226.翻转二叉树[https://leetcode.cn/problems/invert-binary-tree/description/]
递归：递归三部曲：①确定递归函数的参数和返回值 ②确定终止条件 ③确定单层递归的逻辑

/**
 * 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;
        }
        swap(root);
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }

    public void swap(TreeNode p) {
        TreeNode q = p.left;
        p.left = p.right;
        p.right = q;
    }
}

迭代：直接套用迭代前序遍历模板即可，将visit()调整为swap()即可。

/**
 * 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> sc = new Stack<>();
        sc.push(root);
        TreeNode p;
        while (!sc.isEmpty()) {
            while (!sc.isEmpty()) {
                p = sc.pop();
                swap(p);//此处就是visit(p);
                if (p.right != null) {
                    sc.push(p.right);
                }
                if (p.left != null) {
                    sc.push(p.left);
                }
            }
        }
        return root;
    }

    public void swap(TreeNode p) {
        TreeNode q = p.left;
        p.left = p.right;
        p.right = q;
    }
}

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102.二叉树的层序遍历[https://leetcode.cn/problems/binary-tree-level-order-traversal/]
思路：与迭代前序遍历类似，重点在于int loop = que.size();使用loop记录当前数组大小，使用for一次性弹出多个元素出栈记录，入栈非空孩子节点

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

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199.二叉树的右视图[https://leetcode.cn/problems/binary-tree-right-side-view/]
思路：层次遍历二叉树的扩展，重点在于使用loop循环遍历元素，退出循环时，p所指向的元素就是每一层最右边的元素，故