遍历Java中二叉树的所有节点

Question

假设我有一个简单的二叉树节点类,如下所示:

public class BinaryTreeNode {
    public String identifier = "";
    public BinaryTreeNode parent = null;
    public BinaryTreeNode left = null;
    public BinaryTreeNode right = null;

    public BinaryTreeNode(BinaryTreeNode parent, String identifier)
    {
        this.parent = parent; //passing null makes this the root node
        this.identifier = identifier;
    }

    public boolean IsRoot() {
        return parent == null;
    }
}

我如何添加一个能够以递归方式遍历任何大小树的方法,从左到右访问每个现有节点,而无需重新访问已遍历的节点？

这会有用吗？:

public void traverseFrom(BinaryTreeNode rootNode)
{
    /* insert code dealing with this node here */

    if(rootNode.left != null)
        rootNode.left.traverseFrom(rootNode.left);

    if(rootNode.right != null)
        rootNode.traverseFrom(rootNode.right);
}

Answer 1

您可以实现三种类型的二叉树遍历:

例:

考虑以下二叉树:

Pre-order traversal sequence: F, B, A, D, C, E, G, I, H (root, left, right)
In-order traversal sequence: A, B, C, D, E, F, G, H ,I (left, root, right)
Post-order traversal sequence: A, C, E, D, B, H, I, G, F (left, right, root)

代码示例:

从二进制树的左到右遍历,nay为了遍历二叉树:

public void traverse (Node root){ // Each child of a tree is a root of its subtree.
    if (root.left != null){
        traverse (root.left);
    }
    System.out.println(root.data);
    if (root.right != null){
        traverse (root.right);
    }
}

Answer 2

codeMan是对的.遍历将访问左侧的每个节点.一旦它到达左侧的最后一个节点,它就会开始沿着右侧节点返回.这是深度优先搜索(DFS)遍历.因此,每个节点仅被访问一次,并且算法在O(n)时间内运行.快乐的编码.