Outputs all leaf nodes C of the binary tree

Summary. A binary tree is a finite set of n (n 0) nodes:

1), or an empty binary tree, i.e. n = 0.

2), or it consists of a root node and two disjoint left and right subtrees called roots. The left and right subtrees are a binary tree, respectively.

Binary trees only have root nodes, how to represent them by c.

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A binary tree is a finite set of n (n 0) nodes: (1), or an empty binary tree, i.e., n = 0. 2), or it consists of a root node and two disjoint left and right subtrees called roots.

The left and right subtrees are a binary tree, respectively.

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250 leaf nodes.

Let the height be h, because the total number of nodes is 500, greater than 2 8 and less than 2 9; Therefore, the height h should be (8+1)=9, then the bottom layer n9=500-(2 9-1)=245, and the penultimate layer n8=2 7=128; The bottom 245 leaf nodes correspond to 123 parent nodes, so there are 128-123 leaf nodes in the penultimate layer = 5; So the leaf node n=245+5=250

Complete binary tree (complete

binary

tree) if the height of the binary tree is h, divide by the first.

Layer H, other layers.

1～h-1)

The number of nodes has reached the maximum number.

The H layer is continuously missing a number of nodes from right to left, which is a complete binary tree.

Leaf nodes can only appear on the two largest layers, and for any node, if the maximum level of descendants under its right branch is l, the maximum level of its descendants under its left branch must be l

or L+1 binary tree is a very important class of tree structures, which can be recursively defined as follows:

A binary tree t is a set of finite nodes, either as an empty set, or as a root node u and two disjoint binary trees u(1) and u(2) called the left and right subtrees, respectively. If n, n1 and n2 are used to denote the number of nodes of t, u(1) and u(2) respectively, then there is n=1+n1+n2

u(1) and u(2) are sometimes referred to as the first and second subtrees of t, respectively.

Thus, the root of a binary tree can be empty left subtree or empty right subtree, or both left and right subtrees can be empty.

In a binary tree, each node has a maximum of two sons, and there are left and right sons. Thus there are no more than 4 cases of sons at any node: there are no sons; There is only one left son; There is only one right son; There was a left son and there was a right son.

Importance of the binary tree: In any binary tree, the total number of leaf nodes is 1 more than that of a node of 2.

Proof: Let n0 be the number of leaf nodes of the binary tree; n1 is the number of nodes in the binary tree with a moderate degree of 1; n2 is the number of nodes in the binary tree with a moderate degree of 2, and it is clear that n=n0+n1+n2 (1).

Because in the binary tree, except for the root node, each node has and only one antecedent. Let b be the number of predecessors of the binary tree, n=b+1(2).

All of these antecedents are at the same time posteriors to the nodes of degree 1 and degree 2. Thus there is b=n1+2n2 (3).

We substitute (3) for (2) to get n=n1+2n2+1 (4).

Comparing (1) and (4), it is obtained that n0=n2+1, that is, the number of leaves is 1 more than the number of nodes with a degree of 2

It's a theorem.

A binary tree is a tree in which nodes fork no more than two.

That is, the degree of the whole tree is 2

If the first layer (root) is layer 1, there is generally a relationship:

There are the most on the i-layer.

2^(i-1)

nodes i>=1

A binary tree with a depth of k has a maximum of 2 k-1

nodes have. The full binary tree depth of n nodes is .

log 2(n) is rounded down +1

For details, please refer to Yan Weimin's "Data Structure".

The number of nodes with a degree of 0 (i.e. the number of leaf nodes) = the number of nodes with a degree of 2 + 1. In the problem, the number of leaf nodes is 22, and the number of nodes with a degree of 2 can be calculated as 21 by using the property. There are only three types of nodes in a binary tree:

The total number of nodes with degree 0, degree 1, and degree 2 is 25, so the number of nodes with degree 1 is 22 + 13 + 21 = 56.

Because the relationship between the leaf node and the node of degree 2 is: n0 n2 1;

Because n0 22, n2 2;

Total number of nodes: n n0 n1 n2 22 13 2 37

If the root node is hierarchical at 1, then:

A complete binary tree of n nodes, with a depth of rounding down [log2n] +1 or rounding up [log2(n+ 1)], the specific process is found in almost all data structure textbooks, and it is launched by using the nature of binary trees.

The parents of i are numbered by rounding down [i 2], the left child is numbered 2i, and the right child is numbered 2i + 1

All of this can be proved by mathematical induction.

If a binary tree has n2 nodes with degrees of 2, then the binary tree has (n2+1) leaf nodes.

Let n be the total number of nodes answered, n0 is the number of nodes with a degree of 0 (i.e., leaf nodes), n1 is the number of nodes with a degree of 1, and n2 is the number of nodes with a degree of 2, then there is n=n0+n1+n2, n-1=n1+2*n2=total number of branches, and from these two equations, we can get: n0=n2+1.

n2=5, then there are 6 leaves.

When the number of edges is equal to n-1, the number of leaf nodes is (n 2+1), and n is an odd number, the result is rounded (not rounded).

The number of nodes in a full binary tree is 2 to the power of 2 (d-1).

This is the formula that prescribes death.