代码随想录：二叉树层序遍历系列

102、二叉树的层序遍历

/**
 * 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<List<Integer>>() ;
        if(root==null)return ans;
        Queue<TreeNode> q=new LinkedList<TreeNode>();
        q.offer(root);
        while(q.size()>0)
        {
            ArrayList<Integer> cur=new ArrayList<>();
            int n=q.size();
            for(int i=0;i<n;i++)
            {
                TreeNode t=q.peek();
                q.poll();
               cur.add(t.val);
               if(t.left!=null)q.offer(t.left);
               if(t.right!=null)q.offer(t.right);

            }
            ans.add(cur);
        }
        return ans;
    }
}

107、二叉树的层序遍历||

/**
 * 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 {
    List<List<Integer>> ans=new ArrayList<List<Integer>>();
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
      //  List<List<Integer>> ans=new  ArrayList<List<Integer>>() ;
        if(root==null)return ans;
        Queue<TreeNode> q=new LinkedList<TreeNode>();
        q.offer(root);
        while(!q.isEmpty())
        {
            ArrayList<Integer> cur=new ArrayList<>();
            int n=q.size();
            for(int i=0;i<n;i++)
            {
                TreeNode t=q.poll();
               cur.add(t.val);
               if(t.left!=null)q.add(t.left);
               if(t.right!=null)q.add(t.right);

            }
            ans.add(cur);
        }
        Collections.reverse(ans);
        return ans;
    }
}

199. 二叉树的右视图

/**
 * 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 {
    List<Integer> ans=new ArrayList<Integer>();
    public List<Integer> rightSideView(TreeNode root) {
        if(root==null)return ans;
        traval(root,0);
        return ans;
    }
    void traval(TreeNode root,int deep )
    {
        if(root==null)return ;
        deep++;
        if(ans.size()<deep)
        {
            ans.add(null);
        }
        if(ans.get(deep-1)==null)ans.set(deep-1,root.val);
        traval(root.right,deep);
        traval(root.left,deep);
        return ;
    }
}

637. 二叉树的层平均值

/**
 * 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<Double> averageOfLevels(TreeNode root) {
        Queue<TreeNode> q=new LinkedList<TreeNode>();
        q.offer(root);
        List<Double> ans=new ArrayList<>();
        while(!q.isEmpty())
        {
            int n=q.size();
            double sum=0;
            for(int i=0;i<n;i++)
            {
                TreeNode cur=q.poll();
                sum+=cur.val;
                if(cur.left!=null)q.add(cur.left);
                if(cur.right!=null)q.add(cur.right);
            }
             ans.add(sum/n);
        }
        return ans;
    }
}

429. N 叉树的层序遍历

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> ans=new ArrayList<List<Integer>>();
        Queue<Node> q=new LinkedList<>();
        if(root==null)return ans;
        q.add(root);
        while(!q.isEmpty())
        {
            int n=q.size();
            List<Integer> cur=new ArrayList<Integer>();
            for(int i=0;i<n;i++)
           {
            Node t=q.poll();
            cur.add(t.val);
            int l=t.children.size();
            for(int j=0;j<l;j++)
            {
              q.add(t.children.get(j));
            }
           }
           ans.add(cur);
        }
        return ans;
            }
}

515. 在每个树行中找最大值

/**
 * 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<Integer> largestValues(TreeNode root) {
        List<Integer> ans=new ArrayList<>();
        Queue<TreeNode> q=new LinkedList<>();
        if(root==null)return ans;
        q.add(root);
        while(!q.isEmpty())
        {
            int n=q.size();
            List<Integer> cur=new ArrayList<>();
            for(int i=0;i<n;i++)
            {
                TreeNode t=q.poll();
                cur.add(t.val);
                if(t.left!=null)q.add(t.left);
                if(t.right!=null)q.add(t.right);
            }
            ans.add(Collections.max(cur));
        }
        return ans;
            }
}

116. 填充每个节点的下一个右侧节点指针

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}
    
    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/

class Solution {
    public Node connect(Node root) {
        if(root==null)return root;
        Deque<Node> q=new LinkedList<>();
        q.push(root);
        while(!q.isEmpty())
        {
            int n=q.size();
            Node t;
            for(int i=0;i<n-1;i++)
            {
               t=q.poll();
              if(t.left!=null)q.add(t.left);
              if(t.right!=null)q.add(t.right);
               t.next=q.peek();
            }
            t=q.poll();
            if(t.left!=null)q.add(t.left);
              if(t.right!=null)q.add(t.right);
              t.next=null;
        }
        return root;
    }
}

117. 填充每个节点的下一个右侧节点指针 II

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}
    
    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/

class Solution {
    public Node connect(Node root) {
        Queue<Node> q=new LinkedList<>();
        if(root==null)return root;
        q.add(root);
        while(!q.isEmpty())
        {
            int n=q.size();
            Node t;
            for(int i=0;i<n-1;i++)
            {
             t=q.poll();
             if(t.left!=null)q.add(t.left);
             if(t.right!=null)q.add(t.right);
             t.next=q.peek();
            }
            t=q.poll();
             if(t.left!=null)q.add(t.left);
             if(t.right!=null)q.add(t.right);
             t.next=null;
        }
        return root;
    }
}

104. 二叉树的最大深度

/**
 * 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 int maxDepth(TreeNode root) {
        int ans=0;
        Queue<TreeNode> q=new LinkedList<>();
        if(root==null)return ans;
        q.add(root);
        while(!q.isEmpty())
        {
            ans++;
            int n=q.size();
            for(int i=0;i<n;i++)
            {
                TreeNode cur=q.poll();
                if(cur.left!=null)q.add(cur.left);
                if(cur.right!=null)q.add(cur.right);
            } 
        }
        return ans;
    }
}

111. 二叉树的最小深度

/**
 * 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 int minDepth(TreeNode root) {
        int ans=0;
        if(root==null)return ans;
        Queue<TreeNode> q=new LinkedList<>();
        q.add(root);
        while(!q.isEmpty())
        {
            ans++;
            int n=q.size();
            for(int i=0;i<n;i++)
            {
                TreeNode t=q.poll();
                if(t.left==t.right==null)return ans;
                if(t.left!=null)q.add(t.left);
                if(t.right!=null)q.add(t.right);
            }
        }
        return ans;
    }
}