LeetCode —— 二叉树

持续更新中................

二叉树的定义

public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    public TreeNode() {
    }

    public TreeNode(int val) {
        this.val = val;
    }

    public TreeNode(int val, TreeNode left, TreeNode right) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

144. 二叉树的前序遍历

给你二叉树的根节点 root ，返回它节点值的前序遍历。

示例1：输入：root = [1,null,2,3] 输出：[1,2,3]

//递归遍历
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        preorder(root, list);
        return list;
    }

    public void preorder(TreeNode root, List<Integer> list){
        if(root == null){
            return;
        }
        list.add(root.val);
        preorder(root.left, list);
        preorder(root.right, list);
    }
}

//迭代遍历
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if(root == null){
            return list;
        }
        
        // 前序遍历顺序：中-左-右，入栈顺序：中-右-左
        Deque<TreeNode> deque = new LinkedList<>();
        deque.push(root);
        while(!deque.isEmpty()){
            TreeNode node = deque.pop();
            list.add(node.val);
            if(node.right != null){
                deque.push(node.right);
            }
            if(node.left != null){
                deque.push(node.left);
            }
        }
        return list;
    }
}

//迭代遍历（前序、后序、中序的统一形式）
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        Deque<TreeNode> deque = new LinkedList<>();

        if(root != null){
            deque.push(root);
        }

        while(!deque.isEmpty()){
            TreeNode node = deque.peek();
            if(node != null){
                deque.pop();
                //前序遍历顺序：中-左-右，入栈顺序：右-左-中，出栈顺序：中-左-右
                if(node.right != null){
                    deque.push(node.right);
                }
                if(node.left != null){
                    deque.push(node.left);
                }
                deque.push(node);

                deque.push(null);   //加入空节点做为标记，来将中节点取出
            }else{
                deque.pop();    //将空节点弹出
                
                node = deque.peek();
                list.add(node.val);   
                deque.pop();    
            }
        }

        return list;
    }
}

145. 二叉树的后序遍历

给你一棵二叉树的根节点 root ，返回其节点值的 后序遍历 。

示例1：输入：root = [1,null,2,3] 输出：[3,2,1]

//递归遍历
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        postorder(root, list);
        return list;
    }

    public void postorder(TreeNode root, List<Integer> list){
        if(root == null){
            return;
        }
        postorder(root.left, list);
        postorder(root.right, list);
        list.add(root.val);
    }
}

//迭代遍历
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if(root == null){
            return list;
        }

        // 后序遍历顺序 左-右-中 入栈顺序：中-左-右 出栈顺序：中-右-左， 最后翻转结果得到 左-右-中
        Deque<TreeNode> deque = new LinkedList<>();
        deque.push(root);
        while(!deque.isEmpty()){
            TreeNode node = deque.pop();
            list.add(node.val);
            if(node.left != null){
                deque.push(node.left);
            }
            if(node.right != null){
                deque.push(node.right);
            }

        }
        Collections.reverse(list);
        return list;
    }
}

//迭代遍历（前序、后序、中序的统一形式）
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        Deque<TreeNode> deque = new LinkedList<>();

        if(root != null){
            deque.push(root);
        }

        while(!deque.isEmpty()){
            TreeNode node = deque.peek();
            if(node != null){
                deque.pop();

                //后序遍历顺序：左-右-中，入栈顺序：中-右-左，出栈顺序：左-右-中
                deque.push(node);
                deque.push(null);   //加入空节点做为标记，来将节点取出
                if(node.right != null){
                    deque.push(node.right);
                }
                if(node.left != null){
                    deque.push(node.left);
                }
            }else{
                deque.pop();    //将空节点弹出

                node = deque.peek();
                list.add(node.val);
                deque.pop();
            }
        }

        return list;
    }
}

94. 二叉树的中序遍历

给定一个二叉树的根节点 root ，返回 它的中序遍历 。

示例1：输入：root = [1,null,2,3] 输出：[1,3,2]

//递归遍历
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        inorder(root, list);
        return list;
    }

    public void inorder(TreeNode root, List<Integer> list){
        if(root == null){
            return;
        }
        inorder(root.left, list);
        list.add(root.val);
        inorder(root.right, list);
    }
}

//迭代遍历
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if(root == null){
            return list;
        }

        //中序遍历顺序: 左-中-右 入栈顺序： 左-右
        Deque<TreeNode> deque = new LinkedList<>();
        TreeNode node = root;
        while(node != null || !deque.isEmpty()){
            if(node != null){
                deque.push(node);
                node = node.left;
            }else{
                node = deque.pop();
                list.add(node.val);
                node = node.right;
            }
        }
        return list;
    }
}

//迭代遍历（前序、后序、中序的统一形式）
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        Deque<TreeNode> deque = new LinkedList<>();

        if(root != null){
            deque.push(root);
        }

        while(!deque.isEmpty()){
            TreeNode node = deque.peek();
            if(node != null){
                deque.pop();

                //中序遍历顺序：左-中-右，入栈顺序：右-中-左，出栈顺序：左-中-右
                if(node.right != null){
                    deque.push(node.right);
                }
                deque.push(node);
                deque.push(null);   //加入空节点做为标记，来将节点取出
                if(node.left != null){
                    deque.push(node.left);
                }
            }else{
                deque.pop();    //将空节点弹出

                node = deque.peek();
                list.add(node.val);
                deque.pop();
            }
        }

        return list;
    }
}

102. 二叉树的层序遍历

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]]

//递归遍历
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> list = new ArrayList<>();
        levelOrderTraversal(root, list, 0);
        return list;
    }

    public void levelOrderTraversal(TreeNode root, List<List<Integer>> list, Integer deep){
        if(root == null){
            return;
        }

        deep++;

        if(list.size() < deep){
            List<Integer> item = new ArrayList<>();
            list.add(item);
        }
        list.get(deep - 1).add(root.val);

        levelOrderTraversal(root.left, list, deep);
        levelOrderTraversal(root.right, list, deep);
    }
}

//迭代遍历
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> list = new ArrayList<>();

        if(root == null){
            return list;
        }

        Deque<TreeNode> deque = new LinkedList<>();
        deque.offer(root);
        while(!deque.isEmpty()){
            List<Integer> item = new ArrayList<>();
            int len = deque.size();
            for(int i = 0; i < len; i++){
                TreeNode node = deque.poll();
                item.add(node.val);

                if(node.left != null){
                    deque.offer(node.left);
                }
                if(node.right != null){
                    deque.offer(node.right);
                }
            }
            list.add(item);
        }
        return list;
    }
}

107. 二叉树的层序遍历II

给你二叉树的根节点 root ，返回其节点值 自底向上的层序遍历 。（即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

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

//递归遍历
class Solution107 {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> list = new ArrayList<>();
        levelOrderTraversal(root, list, 0);
        Collections.reverse(list);
        return list;
    }

    public void levelOrderTraversal(TreeNode root, List<List<Integer>> list, Integer deep){
        if(root == null){
            return;
        }

        deep++;

        if(list.size() < deep){
            List<Integer> item = new ArrayList<>();
            list.add(item);
        }
        list.get(deep - 1).add(root.val);

        levelOrderTraversal(root.left, list, deep);
        levelOrderTraversal(root.right, list, deep);
    }
}

//迭代遍历
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> list = new ArrayList<>();

        if(root == null){
            return list;
        }

        Deque<TreeNode> deque = new LinkedList<>();
        deque.offer(root);
        while(!deque.isEmpty()){
            List<Integer> item = new ArrayList<>();
            int len = deque.size();
            for(int i = 0; i < len; i++){
                TreeNode node = deque.poll();
                item.add(node.val);

                if(node.left != null){
                    deque.offer(node.left);
                }
                if(node.right != null){
                    deque.offer(node.right);
                }
            }
            list.add(item);
        }

        Collections.reverse(list);
        return list;
    }
}

199. 二叉树的右视图

给定一个二叉树的根节点 root，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。

示例1：输入: [1,2,3,null,5,null,4] 输出: [1,3,4]

//递归遍历
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<>();

        List<List<Integer>> list = new ArrayList<>();
        levelOrderTraversal(root, list, 0);

        //先层序遍历，再取每一层最右侧的值
        for(List<Integer> item : list){
            result.add(item.get(item.size() - 1));
        }
        return result;
    }

    public void levelOrderTraversal(TreeNode root, List<List<Integer>> list, Integer deep){
        if(root == null){
            return;
        }

        deep++;

        if(list.size() < deep){
            List<Integer> item = new ArrayList<>();
            list.add(item);
        }
        list.get(deep - 1).add(root.val);

        levelOrderTraversal(root.left, list, deep);
        levelOrderTraversal(root.right, list, deep);
    }
}

//迭代遍历
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> list = new ArrayList<>();

        if(root == null){
            return list;
        }

        Deque<TreeNode> deque = new LinkedList<>();
        deque.offer(root);
        while(!deque.isEmpty()){
            int len = deque.size();
            for(int i = 0; i < len; i++){
                TreeNode node = deque.poll();
                if(i == len - 1){
                    list.add(node.val);
                }
                if(node.left != null){
                    deque.offer(node.left);
                }
                if(node.right != null){
                    deque.offer(node.right);
                }
            }
        }

        return list;
    }
}

637. 二叉树的层平均值

给定一个非空二叉树的根节点 root , 以数组的形式返回每一层节点的平均值。

示例1： 输入：root = [3,9,20,null,null,15,7] 输出：[3.00000,14.50000,11.00000]

//递归遍历
class Solution637 {
    public List<Double> averageOfLevels(TreeNode root) {
        List<List<Integer>> list = new ArrayList<>();
        levelOrderTraversal(root, list, 0);

        List<Double> result = new ArrayList<>();
        for(List<Integer> item : list){
            double sum = 0;
            for(int i : item){
                sum += i;
            }
            result.add(sum / item.size());
        }

        return result;
    }

    public void levelOrderTraversal(TreeNode root, List<List<Integer>> list, Integer deep){
        if(root == null){
            return;
        }

        deep++;

        if(list.size() < deep){
            List<Integer> item = new ArrayList<>();
            list.add(item);
        }
        list.get(deep - 1).add(root.val);

        levelOrderTraversal(root.left, list, deep);
        levelOrderTraversal(root.right, list, deep);
    }
}

//迭代遍历
class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> list = new ArrayList<>();

        if(root == null){
            return list;
        }

        Deque<TreeNode> deque = new LinkedList<>();
        deque.offer(root);
        while(!deque.isEmpty()){
            double sum = 0;
            int len = deque.size();
            for(int i = 0; i < len; i++){
                TreeNode node = deque.poll();
                sum += node.val;

                if(node.left != null){
                    deque.offer(node.left);
                }
                if(node.right != null){
                    deque.offer(node.right);
                }
            }
            list.add(sum / len);
        }

        return list;
    }
}

429. N叉树的层序遍历

给定一个 N 叉树，返回其节点值的层序遍历。（即从左到右，逐层遍历）。

树的序列化输入是用层序遍历，每组子节点都由 null 值分隔（参见示例）。

示例1：输入：root = [1,null,3,2,4,null,5,6] 输出：[[1],[3,2,4],[5,6]]

class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int val) {
        this.val = val;
    }

    public Node(int val, List<Node> children) {
        this.val = val;
        this.children = children;
    }
}

//递归
class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> list = new ArrayList<>();
        levelOrderTraversal(root, list, 0);
        return list;
    }

    public void levelOrderTraversal(Node root, List<List<Integer>> list, Integer deep){
        if(root == null){
            return;
        }

        deep++;

        if(list.size() < deep){
            List<Integer> item = new ArrayList<>();
            list.add(item);
        }
        list.get(deep - 1).add(root.val);

        for(Node child : root.children){
            levelOrderTraversal(child, list, deep);
        }
    }
}

//迭代
class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> list = new ArrayList<>();
        if(root == null){
            return list;
        }

        Deque<Node> deque = new LinkedList<>();
        deque.offer(root);
        while(!deque.isEmpty()){
            List<Integer> item = new ArrayList<>();
            int len = deque.size();
            for(int i = 0; i < len; i++){
                Node node = deque.poll();
                item.add(node.val);

                for(Node child : node.children){
                    deque.offer(child);
                }
            }
            list.add(item);
        }

        return list;
    }
}

515. 在每个树行中找最大值

给定一棵二叉树的根节点 root ，请找出该二叉树中每一层的最大值。

示例1：输入: root = [1,3,2,5,3,null,9] 输出: [1,3,9]

//递归
class Solution {
    public List<Integer> largestValues(TreeNode root) {
        List<Integer> result = new ArrayList<>();

        List<List<Integer>> list = new ArrayList<>();
        levelOrderTraversal(root, list, 0);
        
        for(List<Integer> item : list){
            result.add(Collections.max(item));
        }
        return result;
    }

    public void levelOrderTraversal(TreeNode root, List<List<Integer>> list, int deep){
        if(root == null){
            return;
        }
        
        deep++;
        
        if(list.size() < deep){
            List<Integer> item = new ArrayList<>();
            list.add(item);
        }
        list.get(deep - 1).add(root.val);

        levelOrderTraversal(root.left, list, deep);
        levelOrderTraversal(root.right, list, deep);
    }
}

//迭代
class Solution {
    public List<Integer> largestValues(TreeNode root) {
        List<Integer> result = new ArrayList<>();

        if(root == null){
            return result;
        }

        Deque<TreeNode> deque = new LinkedList<>();
        deque.offer(root);
        while(!deque.isEmpty()){
            int maxValue = Integer.MIN_VALUE;
            int len = deque.size();
            for(int i = 0; i < len; i++){
                TreeNode node = deque.poll();
                maxValue = node.val > maxValue? node.val : maxValue;

                if(node.left != null){
                    deque.offer(node.left);
                }
                if(node.right != null){
                    deque.offer(node.right);
                }
            }
            result.add(maxValue);
        }

        return result;
    }
}

116. 填充每个节点的下一个右侧节点指针

给定一个 完美二叉树 ，其所有叶子节点都在同一层，每个父节点都有两个子节点。填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。

示例1：输入：root = [1,2,3,4,5,6,7] 输出：[1,#,2,3,#,4,5,6,7,#]

class Solution {
    public Node connect(Node root) {
        if(root == null){
            return null;
        }

        Deque<Node> deque = new LinkedList<>();
        deque.offer(root);
        while(!deque.isEmpty()){
            int len = deque.size();
            for(int i = 0; i < len; i++){
                Node node = deque.poll();
                if(i < len - 1){
                    node.next = deque.peek();
                }

                if(node.left != null){
                    deque.offer(node.left);
                }
                if(node.right != null){
                    deque.offer(node.right);
                }
            }
        }
        return root;
    }
}

public class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}

    public Node(int val) {
        this.val = val;
    }

    public Node(int val, Node left, Node right, Node next) {
        this.val = val;
        this.left = left;
        this.right = right;
        this.next = next;
    }
}

117. 填充每个节点的下一个右侧节点指针II

给定一个二叉树，填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。初始状态下，所有 next 指针都被设置为 NULL。

示例1：输入：root = [1,2,3,4,5,null,7] 输出：[1,#,2,3,#,4,5,7,#]

class Solution {
    public Node connect(Node root) {
        if(root == null){
            return null;
        }

        Deque<Node> deque = new LinkedList<>();
        deque.offer(root);
        while(!deque.isEmpty()){
            int len = deque.size();
            for(int i = 0; i < len; i++){
                Node node = deque.poll();
                if(i < len - 1){
                    node.next = deque.peek();
                }

                if(node.left != null){
                    deque.offer(node.left);
                }
                if(node.right != null){
                    deque.offer(node.right);
                }
            }
        }
        return root;
    }
}