A tree is a nonlinear data structure, which consists of n ( n>=0 ) finite nodes to form a set with hierarchical relationship. It is called a tree because it looks like an upside-down tree, that is, it has the roots up and the leaves down .

1.2 Concept (important)

a. The degree of the node: the number of subtrees of the node; as shown in the figure above: the degree of A is 6, and the degree of J is 2

b. The degree of the tree: in the tree, the degree of the largest node is the degree of the number; as shown above: the degree of the tree is 6

c. Leaf nodes (terminal nodes) : nodes with degree 0 (nodes without subtrees)

d. Parent node/parent node: as shown above: D is the parent node of H

Child node/child node : As shown above: H is the child node of D

e. Root node: a node without parents; as shown above: A

f. The level of nodes : starting from the definition of the root, the root is the first layer , the child nodes of the root are the second layer, and so on;

g. The height or depth of the tree: the maximum level of nodes in the tree; as shown above: the height of the tree is 4

2. Binary tree (emphasis)

2.1 Concept

Each node has at most two subtrees, degree <= 2.

2.2 The basic form of binary tree

2.3 Two special binary trees

a. Full binary tree: non-cotyledon degree is 2

b. Complete binary tree: The full binary tree is missing the "lower right corner"

2.4 Properties of Binary Trees

a. Full binary tree:

1. If the height is K, there are 2^k-1 nodes

2. If the level is K, then the level has 2^(k-1) nodes

3. Number of Edges = Number of Nodes - 1

4. If the degree is 0, there are n0, and if the degree is 2, there are n2, then n0 = n2 + 1

b. Complete binary tree:

1. If there is a right child, there must be a left child

2. There can only be one node with degree 1

2.5 Storage of binary tree

The storage structure of binary tree is divided into: sequential storage and linked storage similar to linked list .

Sequential storage: only complete binary trees can be stored

Chained Storage: Ordinary Binary Tree

This show chain storage

The chain storage of the binary tree is referenced through one node, and the common representations include binary and ternary representations .

Take this picture as an example, as follows:

// 孩子表示法
private static class TreeNode{
    char val;
    TreeNode left;
    TreeNode right;

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

initialization:

    public static TreeNode build(){
        TreeNode nodeA=new TreeNode('A');
        TreeNode nodeB=new TreeNode('B');
        TreeNode nodeC=new TreeNode('C');
        TreeNode nodeD=new TreeNode('D');
        TreeNode nodeE=new TreeNode('E');
        TreeNode nodeF=new TreeNode('F');
        TreeNode nodeG=new TreeNode('G');
        TreeNode nodeH=new TreeNode('H');
        nodeA.left=nodeB;
        nodeA.right=nodeC;
        nodeB.left=nodeD;
        nodeB.right=nodeE;
        nodeE.right=nodeH;
        nodeC.left=nodeF;
        nodeC.right=nodeG;
        return nodeA;
    }

2.6 Basic operations of binary trees

2.6.1 Binary tree traversal (recursion)

1. NLR : Preorder traversal (Preorder Traversal also known as preorder traversal ) - visit the root node ---> the left subtree of the root ---> the right subtree of the root.

    //先序遍历 : 根左右
    public static void preOrder(TreeNode root){
        if(root==null){
            return;
        }
        System.out.print(root.val+" ");
        preOrder(root.left);
        preOrder(root.right);
    }

2. LNR : Inorder Traversal - left subtree of the root --- > root node ---> right subtree of the root.

    //中序遍历
    public static void inOrder(TreeNode root){
        if(root==null){
            return;
        }
        preOrder(root.left);
        System.out.print(root.val+" ");
        preOrder(root.right);
    }

3. LRN : Postorder Traversal - left subtree of the root ---> right subtree of the root ---> root node.

    //后序遍历
    public static void postOrder(TreeNode root){
        if(root==null){
            return;
        }
        preOrder(root.left);
        preOrder(root.right);
        System.out.print(root.val+" ");
    }

2.6.2 Traversal (iteration) of binary tree

1. Preorder traversal

    //方法2(迭代)
    //先序遍历 (迭代)
    public static void preOrderNonRecursion(TreeNode root){
        if(root==null){
            return ;
        }
        Deque<TreeNode> stack=new LinkedList<>();
        stack.push(root);
        while (!stack.isEmpty()){
            TreeNode cur=stack.pop();
            System.out.print(cur.val+" ");
            if(cur.right!=null){
                stack.push(cur.right);
            }
            if(cur.left!=null){
                stack.push(cur.left);
            }
        }
    }

2. Inorder traversal

    //方法2(迭代)
    //中序遍历 (迭代)
    public static void inorderTraversalNonRecursion(TreeNode root) {
        if(root==null){
            return ;
        }

        Deque<TreeNode> stack=new LinkedList<>();
        // 当前走到的节点
        TreeNode cur=root;
        while (!stack.isEmpty() || cur!=null){
            // 不管三七二十一，先一路向左走到根儿~
            while (cur!=null){
                stack.push(cur);
                cur=cur.left;
            }
            // 此时cur为空，说明走到了null，此时栈顶就存放了左树为空的节点
            cur=stack.pop();
            System.out.print(cur.val+" ");
            // 继续访问右子树
            cur=cur.right;
        }
    }

3. Post-order traversal

    //方法2(迭代)
    //后序遍历 (迭代)
    public static void postOrderNonRecursion(TreeNode root){
        if(root==null){
            return;
        }
        Deque<TreeNode> stack=new LinkedList<>();
        TreeNode cur=root;
        TreeNode prev=null;

        while (!stack.isEmpty() || cur!=null){
            while (cur!=null){
                stack.push(cur);
                cur=cur.left;
            }

            cur=stack.pop();
            if(cur.right==null || prev==cur.right){
                System.out.print(cur.val+" ");
                prev=cur;
                cur=null;
            }else {
                stack.push(cur);
                cur=cur.right;
            }
        }
    }

2.6.3 Basic operations of binary trees

1. Find the number of nodes (recursion & iteration)

    //方法1(递归)
    //传入一颗二叉树的根节点，就能统计出当前二叉树中一共有多少个节点，返回节点数
    //此时的访问就不再是输出节点值，而是计数器 + 1操作
    public static int getNodes(TreeNode root){
        if(root==null){
            return 0;
        }
        return 1+getNodes(root.left)+getNodes(root.right);
    }

    //方法2(迭代)
    //使用层序遍历来统计当前树中的节点个数
    public static int getNodesNoRecursion(TreeNode root){
        if(root==null){
            return 0;
        }
        int size=0;
        Deque<TreeNode> queue=new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            size++;
            if (cur.left != null) {
                queue.offer(cur.left);
            }
            if (cur.right != null) {
                queue.offer(cur.right);
            }
        }
        return size;
    }

2. Find the number of leaf nodes (recursion & iteration)

    //方法1(递归)
    //传入一颗二叉树的根节点，就能统计出当前二叉树的叶子结点个数
    public static int getLeafNodes(TreeNode root){
        if(root==null){
            return 0;
        }
        if(root.left==null && root.right==null){
            return 1;
        }
        return getLeafNodes(root.left)+getLeafNodes(root.right);
    }

    //方法2(迭代)
    //使用层序遍历来统计叶子结点的个数
    public static int getLeafNodesNoRecursion(TreeNode root){
        if(root==null){
            return 0;
        }
        int size=0;
        Deque<TreeNode> queue=new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()){
            TreeNode cur=queue.poll();
            if(cur.left==null && cur.right==null){
                size++;
            }
            if(cur.left!=null){
                queue.offer(cur.left);
            }
            if(cur.right!=null){
                queue.offer(cur.right);
            }
        }
        return size;
    }

3. Find the number of nodes in the kth layer

    //求出以root为根节点的二叉树第k层的节点个数
    public static int getKLevelNodes(TreeNode root,int k){
        if(root==null || k<=0){
            return 0;
        }
        if(k==1){
            return 1;
        }
        return getKLevelNodes(root.left,k-1)+getKLevelNodes(root.right,k-1);
    }

4. Find the height of the tree

    //传入一个以root为根节点的二叉树，就能求出该树的高度
    public static int height(TreeNode root){
        if(root==null){
            return 0;
        }
        return 1+ Math.max(height(root.left),height(root.right));
    }

5. Determine whether there is a node with value in the binary tree number

    //判断当前以root为根节点的二叉树中是否包含指定元素val，
    //若存在返回true，不存在返回false
    public static boolean contains(TreeNode root,char value){
        if(root==null){
            return false;
        }
        if(root.val==value){
            return true;
        }
        return contains(root.left,value) || contains(root.right,value);
    }

2.7 Level order traversal of binary tree

    //层序遍历
    public static void levelOrder(TreeNode root) {
        if(root==null){
            return ;
        }

        // 借助队列来实现遍历过程
        Deque<TreeNode> queue =new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()){
            int size=queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur=queue.poll();
                System.out.print(cur.val+" ");
                if(cur.left!=null){
                    queue.offer(cur.left);
                }
                if(cur.right!=null){
                    queue.offer(cur.right);
                }
            }
        }
    }

3. Binary tree complete code

See the following section for the complete code of the binary tree: 《Java binary tree complete code (recursion & iteration) 》

Java binary tree basic concepts (recursion & iteration)

1. Tree structure

1.1 Concept