查找二叉树的所有节点总和的Java程序
1 简介
在此程序中,我们需要计算二叉树中存在的节点总数。首先,我们将遍历左侧子树并计算左侧子树中存在的节点总数。同样,我们计算右子树中存在的节点的总和,并通过添加根的数据来计算总和。
对于给定的树,二叉树的节点总数为1 + 2 + 5 + 8 + 6 + 9 = 31。
2 算法思路
- 定义具有三个属性的Node类,即:左和右数据。在此,左代表节点的左子节点,右代表节点的右子节点。
- 创建节点时,数据将传递到该节点的data属性,并且左右都将设置为null。
- 定义另一个具有属性根的类。
- 根代表树的根节点,并将其初始化为null。
- 它检查根是否为null,这表示树为空。
- 如果树不为空,请遍历左子树,计算节点的总和并将其存储在sumLeft中。
- 然后,遍历右边的子树,计算节点的总和并将其存储在sumRight中。
- 最后,计算总和= temp.data + sumLeft + sumRight。
3 程序实现
/**
* 一点教程网: http://www.yiidian.com
*/
public class SumOfNodes {
//Represent the node of binary tree
public static class Node{
int data;
Node left;
Node right;
public Node(int data){
//Assign data to the new node, set left and right children to null
this.data = data;
this.left = null;
this.right = null;
}
}
//Represent the root of binary tree
public Node root;
public SumOfNodes(){
root = null;
}
//calculateSum() will calculate the sum of all the nodes present in the binary tree
public int calculateSum(Node temp){
int sum, sumLeft, sumRight;
sum = sumRight = sumLeft = 0;
//Check whether tree is empty
if(root == null) {
System.out.println("Tree is empty");
return 0;
}
else {
//Calculate the sum of nodes present in left subtree
if(temp.left != null)
sumLeft = calculateSum(temp.left);
//Calculate the sum of nodes present in right subtree
if(temp.right != null)
sumRight = calculateSum(temp.right);
//Calculate the sum of all nodes by adding sumLeft, sumRight and root node's data
sum = temp.data + sumLeft + sumRight;
return sum;
}
}
public static void main(String[] args) {
SumOfNodes bt = new SumOfNodes();
//Add nodes to the binary tree
bt.root = new Node(5);
bt.root.left = new Node(2);
bt.root.right = new Node(9);
bt.root.left.left = new Node(1);
bt.root.right.left = new Node(8);
bt.root.right.right = new Node(6);
//Display the sum of all the nodes in the given binary tree
System.out.println("Sum of all nodes of binary tree: " + bt.calculateSum(bt.root));
}
}
输出结果为:
um of all nodes of binary tree: 31
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