代码随想录四刷day13

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前言

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学会二叉树的层序遍历，可以一口气打完以下十题

一、力扣144. 二叉树的前序遍历’

/**
 * 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<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<>();
        if(root == null){
            return res;
        }
        Deque<TreeNode> deq = new LinkedList<>();
        TreeNode p = root;
        while(!deq.isEmpty() || p != null){
            if(p != null){
                res.add(p.val);
                deq.offerLast(p);
                p = p.left;
            }else{
                p = deq.pollLast();
                p = p.right;
            }
        }
        return res;
    }
}

二、力扣94. 二叉树的中序遍历

/**
 * 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<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<>();
        Deque<TreeNode> deq = new LinkedList<>();
        if(root == null){
            return res;
        }
        TreeNode p = root;
        while(!deq.isEmpty() || p != null){
            if(p != null){
                deq.offerLast(p);
                p = p.left;
            }else{
                p = deq.pollLast();
                res.add(p.val);
                p = p.right;
            }
        }
        return res;
    }
}

三、力扣145. 二叉树的后序遍历

/**
 * 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<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<>();
        if(root == null){
            return res;
        }
        Deque<TreeNode> deq = new LinkedList<>();
        deq.offerLast(root);
        while(!deq.isEmpty()){
            TreeNode cur = deq.pollLast();
            res.add(cur.val);
            if(cur.left != null){
                deq.offerLast(cur.left);
            }
            if(cur.right != null){
                deq.offerLast(cur.right);
            }
        }
        Collections.reverse(res);
        return res;
    }
}

四、力扣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;3
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        Deque<TreeNode> deq = new LinkedList<>();
        List<List<Integer>> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        deq.offerLast(root);
        while(!deq.isEmpty()){
            int size = deq.size();
            List<Integer> list = new ArrayList<>();
            for(int i = 0; i < size; i ++){
                TreeNode cur = deq.pollFirst();
                list.add(cur.val);
                if(cur.left != null){
                    deq.offerLast(cur.left);
                }
                if(cur.right != null){
                    deq.offerLast(cur.right);
                }
            }
            res.add(list);
        }
        return res;
    }
}

五、力扣107. 二叉树的层序遍历 II

/**
 * 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>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        Deque<TreeNode> deq = new LinkedList<>();
        if(root == null){
            return res;
        }
        deq.offerLast(root);
        while(!deq.isEmpty()){
            int size = deq.size();
            List<Integer> list = new ArrayList<>();
            for(int i = 0; i < size;i ++){
                TreeNode cur = deq.pollFirst();
                list.add(cur.val);
                if(cur.left != null){
                    deq.offerLast(cur.left);
                }
                if(cur.right != null){
                    deq.offerLast(cur.right);
                }
            }
            res.add(list);
        }
        Collections.reverse(res);
        return res;
    }
}

六、力扣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 {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        Deque<TreeNode> deq = new LinkedList<>();
        deq.offerLast(root);
        while(!deq.isEmpty()){
            int size = deq.size();
            for(int i = 0; i < size; i ++){
                TreeNode cur = deq.pollFirst();
                if(i == size-1){
                    res.add(cur.val);
                }
                if(cur.left != null){
                    deq.offerLast(cur.left);
                }
                if(cur.right != null){
                    deq.offerLast(cur.right);
                }
            }
        }
        return res;
    }
}

七、力扣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) {
        List<Double> res = new ArrayList<>();
        Deque<TreeNode> deq = new LinkedList<>();
        deq.offerLast(root);
        while(!deq.isEmpty()){
            double size = deq.size();
            double sum = 0;
            for(double i = 0; i < size; i ++){
                TreeNode cur = deq.pollFirst();
                sum += cur.val;
                if(cur.left != null){
                    deq.offerLast(cur.left);
                }
                if(cur.right != null){
                    deq.offerLast(cur.right);
                }
            }
            res.add(sum/size);
        }
        return res;
    }
}

八、力扣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>> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        Deque<Node> deq = new LinkedList<>();
        deq.offerLast(root);
        while(!deq.isEmpty()){
            int size = deq.size();
            List<Integer> list = new ArrayList<>();
            for(int i = 0; i < size; i ++){
                Node cur = deq.pollFirst();
                list.add(cur.val);
                for(Node node : cur.children){
                    if(node != null){
                        deq.offerLast(node);
                    }
                }
            }
            res.add(list);
        }
        return res;
    }
}

九、力扣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> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        Deque<TreeNode> deq = new LinkedList<>();
        deq.offerLast(root);
        while(!deq.isEmpty()){
            int max = Integer.MIN_VALUE;
            int size = deq.size();
            for(int i = 0; i < size; i ++){
                TreeNode cur = deq.pollFirst();
                max = max > cur.val ? max : cur.val;
                if(cur.left != null){
                    deq.offerLast(cur.left);
                }
                if(cur.right != null){
                    deq.offerLast(cur.right);
                }
            }
            res.add(max);
        }
        return res;
    }
}

十、力扣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) {
        Deque<Node> deq = new LinkedList<>();
        if(root == null){
            return root;
        }
        deq.offerLast(root);
        while(!deq.isEmpty()){
            int size = deq.size();
            Node pre = deq.peekFirst();
            for(int i = 0; i <size; i ++){
                Node cur = deq.pollFirst();
                if(i > 0){
                    pre.next = cur;
                    pre = cur;
                }
                if(cur.left != null){
                    deq.offerLast(cur.left);
                }
                if(cur.right != null){
                    deq.offerLast(cur.right);
                }
            }
        }
        return root;
    }
}

十一、力扣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) {
        if(root == null){
            return root;
        }
        Deque<Node> deq = new LinkedList<>();
        deq.offerLast(root);
        while(!deq.isEmpty()){
            int size = deq.size();
            Node pre = deq.peekFirst();
            for(int i = 0; i < size; i ++){
                Node cur = deq.pollFirst();
                if(i > 0){
                    pre.next = cur;
                    pre = cur;
                }
                if(cur.left != null){
                    deq.offerLast(cur.left);
                }
                if(cur.right != null){
                    deq.offerLast(cur.right);
                }
            }
        }
        return root;
    }
}

十二、力扣04. 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) {
        if(root == null){
            return 0;
        }
        int res = 0;
        Deque<TreeNode> deq = new LinkedList<>();
        deq.offerLast(root);
        while(!deq.isEmpty()){
            res ++;
            int size = deq.size();
            for(int i = 0; i < size; i ++){
                TreeNode cur = deq.pollFirst();
                if(cur.left != null){
                    deq.offerLast(cur.left);
                }
                if(cur.right != null){
                    deq.offerLast(cur.right);
                }
            }
        }
        return res;
    }
}

/**
 * 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) {
        return fun(root);
    }
    public int fun(TreeNode root){
        if(root == null){
            return 0;
        }
        int l = fun(root.left);
        int r = fun(root.right);
        return l > r ? l + 1 : r + 1;
    }
}

十三、力扣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) {
        if(root == null){
            return 0;
        }
        int res = 0;
        Deque<TreeNode> deq = new LinkedList<>();
        deq.offerLast(root);
        while(!deq.isEmpty()){
            res ++;
            int size = deq.size();
            for(int i = 0; i < size; i ++){
                TreeNode cur = deq.pollFirst();
                if(cur.left == null && cur.right == null){
                    return res;
                }
                if(cur.left != null){
                    deq.offerLast(cur.left);
                }
                if(cur.right != null){
                    deq.offerLast(cur.right);
                }
            }
        }
        return res;
    }
}