LeetCode每日三题（七）二叉树.2

一、对称二叉树

自己答案：

不会

标准答案：

/**
 * 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) {
        if(root!=null){
            //如果根结点为空则对称
            //不为空则 左树的左孩子==右数的右孩子
            //左树的右孩子==右数的左孩子
            return yes(root.left,root.right);
        }else
            return true;
    }

    public boolean yes(TreeNode lefttree,TreeNode righttree){
        if(lefttree==null&&righttree==null)
        return true;
        else if(lefttree==null||righttree==null)
        return false;
        return (lefttree.val==righttree.val&&yes(lefttree.left,righttree.right)&&yes(lefttree.right,righttree.left));
    }
}

class Solution {
    public boolean isSymmetric(TreeNode root) {
        return check(root, root);
    }

    public boolean check(TreeNode u, TreeNode v) {
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.offer(u);
        q.offer(v);
        while (!q.isEmpty()) {
            u = q.poll();
            v = q.poll();
            if (u == null && v == null) {
                continue;
            }
            if ((u == null || v == null) || (u.val != v.val)) {
                return false;
            }

            q.offer(u.left);
            q.offer(v.right);

            q.offer(u.right);
            q.offer(v.left);
        }
        return true;
    }
}

二、二叉树的直径、

自己答案：

不会

标准答案：

/**
 * 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 {
    int ans;
    public int depth(TreeNode node) {
        if(node==null){
            return 0;
        }else {
            int left=depth(node.left);
            int right=depth(node.right);
            ans=((left+right+1)>ans)?(left+right+1):ans;
            return (left>right?left:right)+1;
        }
    }

    public  int diameterOfBinaryTree(TreeNode root) {
        ans = 1;
        depth(root);
        return ans - 1;
    }
}

三、二叉树的层序遍历

自己答案：

/**
 * 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) {
       if(root==null){
            List<List<Integer>> list = new ArrayList<>();
            return list;
        }else {
            List<List<Integer>> list = new ArrayList<>();
            Queue<TreeNode> queue = new LinkedList<>();
            queue.offer(root);
            List<Integer> l = new ArrayList<>();
            l.add(root.val);
            list.add(l);

            while (!queue.isEmpty()) {
                //队列不为空
                int size = queue.size();
                List<Integer> temp = new ArrayList<>();
                while (size > 0) {
                    //弹出队列头
                    TreeNode node = queue.remove();
                    if (node.left != null) {
                        temp.add(node.left.val);
                        queue.add(node.left);
                    }
                    if (node.right != null) {
                        temp.add(node.right.val);
                        queue.add(node.right);
                    }
                    size--;
                }
                if(temp.size()!=0){
                    list.add(temp);
                }
            }
            return list;
        }
    }
}

标准答案：

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> ret = new ArrayList<List<Integer>>();
        if (root == null) {
            return ret;
        }

        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            List<Integer> level = new ArrayList<Integer>();
            int currentLevelSize = queue.size();
            for (int i = 1; i <= currentLevelSize; ++i) {
                TreeNode node = queue.poll();
                level.add(node.val);
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            ret.add(level);
        }
        
        return ret;
    }
}