如何获取树的所有叶节点？

Question

假设我在树中有一个节点，如何获取其祖先是该节点的所有叶节点？ 我已经像这样定义了 TreeNode：

public class TreeNode<T>
{
    /** all children of the node */
    private List<TreeNode<T>> children = new ArrayList<TreeNode<T>>();
    /** the parent of the node, if the node is root, parent = null */
    private TreeNode<T> parent = null;
    /** the stored data of the node */
    private T data = null;

    /** the method I want to implement */
    public Set<TreeNode<T>> getAllLeafNodes()
    {
        Set<TreeNode<T>> leafNodes = new HashSet<TreeNode<T>>();
        return leafNodes;
    }
}

Answer 1

使用递归。

• 如果节点本身是叶子，则返回它
• 否则，返回其子节点的所有叶节点

像这样的东西（未测试）：

public Set<TreeNode<T>> getAllLeafNodes() {
    Set<TreeNode<T>> leafNodes = new HashSet<TreeNode<T>>();
    if (this.children.isEmpty()) {
        leafNodes.add(this);
    } else {
        for (TreeNode<T> child : this.children) {
            leafNodes.addAll(child.getAllLeafNodes());
        }
    }
    return leafNodes;
}

Answer 2

创建堆栈并推送根节点。

Stack<Node> st = new Stack<>();
st.push(root);      
CL(st.peek());

调用递归方法。

public void CL(Node root){
        if (st.peek().left == null && st.peek().right == null ) {//if leaf node
            System.out.println(st.peek().data);//print
            st.pop();
            return;
        }
        else{
            if(st.peek().left != null){
                st.add(st.peek().left);
                CL(st.peek());
            }
            if(st.peek().right != null){
                st.add(st.peek().right);
                CL(st.peek());
            }
        }
        st.pop();
    }