当二叉树的节点成为叶节点时打印它们

原文:https://www . geesforgeks . org/print-二叉树的节点成为叶子节点/

给定一棵二叉树。首先打印所有的叶节点，然后从树中移除所有的叶节点，现在打印所有新形成的叶节点，并一直这样做，直到所有的节点都从树中移除。
例:

Input :
8
/ \
3 10
/ \ / \
1 6 14 4
/ \
7 13
Output :
4 6 7 13 14
1 10
3
8


来源 : Flipkart 校园招聘
途径:思路是进行简单的 dfs，根据以下条件给每个节点分配不同的值:

1. 最初将值为 0 的所有节点赋值。


2. 现在，给所有节点赋值为(两个子节点的最大值)+1 。

DFS 之前的树:临时值零被分配给所有节点。

DFS 后的树:节点赋值为(两个子节点的最大值)+1 。

现在，您可以在上面的树中看到，在将所有值分配给每个节点后，任务现在减少到根据分配给它们的节点值的升序来打印树。
以下是上述方法的实施:

C++

// C++ program to print the nodes of binary
// tree as they become the leaf node
#include
using namespace std;
// Binary tree node
struct Node {
    int data;
    int order;
    struct Node* left;
    struct Node* right;
};
// Utility function to allocate a new node
struct Node* newNode(int data, int order)
{
    struct Node* node = new Node;
    node->data = data;
    node->order = order;
    node->left = NULL;
    node->right = NULL;
    return (node);
}
// Function for postorder traversal of tree and
// assigning values to nodes
void Postorder(struct Node* node, vector >& v)
{
    if (node == NULL)
        return;
    /* first recur on left child */
    Postorder(node->left, v);
    /* now recur on right child */
    Postorder(node->right, v);
    // If current node is leaf node, it's order will be 1
    if (node->right == NULL && node->left == NULL) {
        node->order = 1;
        // make pair of assigned value and tree value
        v.push_back(make_pair(node->order, node->data));
    }
    else {
        // otherwise, the order will be:
        // max(left_child_order, right_child_order) + 1
        node->order = max((node->left)->order, (node->right)->order) + 1;
        // make pair of assigned value and tree value
        v.push_back(make_pair(node->order, node->data));
    }
}
// Function to print leaf nodes in
// the given order
void printLeafNodes(int n, vector >& v)
{
    // Sort the vector in increasing order of
    // assigned node values
    sort(v.begin(), v.end());
    for (int i = 0; i         if (v[i].first == v[i + 1].first)
            cout <        else
            cout <    }
}
// Driver Code
int main()
{
    struct Node* root = newNode(8, 0);
    root->left = newNode(3, 0);
    root->right = newNode(10, 0);
    root->left->left = newNode(1, 0);
    root->left->right = newNode(6, 0);
    root->right->left = newNode(14, 0);
    root->right->right = newNode(4, 0);
    root->left->left->left = newNode(7, 0);
    root->left->left->right = newNode(13, 0);
    int n = 9;
    vector > v;
    Postorder(root, v);
    printLeafNodes(n, v);
    return 0;
}


Java 语言(一种计算机语言，尤用于创建网站)

// Java program to print the nodes of binary
// tree as they become the leaf node
import java.util.*;
class GFG
{
// Binary tree node
static class Node
{
    int data;
    int order;
    Node left;
    Node right;
};
static class Pair
{
    int first,second;
    Pair(int a,int b)
    {
        first = a;
        secOnd= b;
    }
}
// Utility function to allocate a new node
static Node newNode(int data, int order)
{
    Node node = new Node();
    node.data = data;
    node.order = order;
    node.left = null;
    node.right = null;
    return (node);
}
static Vector v = new Vector();
// Function for postorder traversal of tree and
// assigning values to nodes
static void Postorder(Node node)
{
    if (node == null)
        return;
    /* first recur on left child */
    Postorder(node.left);
    /* now recur on right child */
    Postorder(node.right);
    // If current node is leaf node, it's order will be 1
    if (node.right == null && node.left == null)
    {
        node.order = 1;
        // make pair of assigned value and tree value
        v.add(new Pair(node.order, node.data));
    }
    else
    {
        // otherwise, the order will be:
        // max(left_child_order, right_child_order) + 1
        node.order = Math.max((node.left).order, (node.right).order) + 1;
        // make pair of assigned value and tree value
        v.add(new Pair(node.order, node.data));
    }
}
static class Sort implements Comparator
{
    // Used for sorting in ascending order of
    // roll number
    public int compare(Pair a, Pair b)
    {
        if(a.first != b.first)
        return (a.first - b.first);
        else
        return (a.second-b.second);
    }
}
// Function to print leaf nodes in
// the given order
static void printLeafNodes(int n)
{
    // Sort the vector in increasing order of
    // assigned node values
    Collections.sort(v,new Sort());
    for (int i = 0; i     {
        if (i != v.size()-1 && v.get(i).first == v.get(i + 1).first)
            System.out.print( v.get(i).second + " ");
        else
            System.out.print( v.get(i).second + "\n");
    }
}
// Driver Code
public static void main(String args[])
{
    Node root = newNode(8, 0);
    root.left = newNode(3, 0);
    root.right = newNode(10, 0);
    root.left.left = newNode(1, 0);
    root.left.right = newNode(6, 0);
    root.right.left = newNode(14, 0);
    root.right.right = newNode(4, 0);
    root.left.left.left = newNode(7, 0);
    root.left.left.right = newNode(13, 0);
    int n = 9;
    Postorder(root);
    printLeafNodes(n);
}
}
// This code is contributed by Arnab Kundu


Python 3

# Python3 program to print the nodes of binary
# tree as they become the leaf node
# Binary tree node
class newNode:
    def __init__(self, data,order):
        self.data = data
        self.order=order
        self.left = None
        self.right = None
# Function for postorder traversal of tree and
# assigning values to nodes
def Postorder(node,v):
    if (node == None):
        return
    """ first recur on left child """
    Postorder(node.left, v)
    """ now recur on right child """
    Postorder(node.right, v)
    # If current node is leaf node,
    # it's order will be 1
    if (node.right == None and
        node.left == None):
        node.order = 1
        # make pair of assigned value and tree value
        v[0].append([node.order, node.data])
    else:
        # otherwise, the order will be:
        # max(left_child_order, right_child_order) + 1
        node.order = max((node.left).order,
                         (node.right).order) + 1
        # make pair of assigned value and tree value
        v[0].append([node.order, node.data])
# Function to print leaf nodes in
# the given order
def printLeafNodes(n, v):
    # Sort the vector in increasing order of
    # assigned node values
    v=sorted(v[0])
    for i in range(n - 1):
        if (v[i][0]== v[i + 1][0]):
            print(v[i][1], end = " ")
        else:
            print(v[i][1])
    if (v[-1][0]== v[-2][0]):
            print(v[-1][1], end = " ")
    else:
        print(v[-1][1])
# Driver Code
root = newNode(8, 0)
root.left = newNode(3, 0)
root.right = newNode(10, 0)
root.left.left = newNode(1, 0)
root.left.right = newNode(6, 0)
root.right.left = newNode(14, 0)
root.right.right = newNode(4, 0)
root.left.left.left = newNode(7, 0)
root.left.left.right = newNode(13, 0)
n = 9
v = [[] for i in range(1)]
Postorder(root, v)
printLeafNodes(n, v)
# This code is contributed by SHUBHAMSINGH10


C

// C# program to print the nodes of binary
// tree as they become the leaf node
using System;
using System.Collections.Generic;
class GFG
{
// Binary tree node
public class Node
{
    public int data;
    public int order;
    public Node left;
    public Node right;
};
public class Pair
{
    public int first,second;
    public Pair(int a,int b)
    {
        first = a;
        secOnd= b;
    }
}
// Utility function to allocate a new node
static Node newNode(int data, int order)
{
    Node node = new Node();
    node.data = data;
    node.order = order;
    node.left = null;
    node.right = null;
    return (node);
}
static List v = new List();
// Function for postorder traversal of
// tree and assigning values to nodes
static void Postorder(Node node)
{
    if (node == null)
        return;
    /* first recur on left child */
    Postorder(node.left);
    /* now recur on right child */
    Postorder(node.right);
    // If current node is leaf node,
    // it's order will be 1
    if (node.right == null &&
        node.left == null)
    {
        node.order = 1;
        // make pair of assigned value
        // and tree value
        v.Add(new Pair(node.order, node.data));
    }
    else
    {
        // otherwise, the order will be:
        // Max(left_child_order,
        //     right_child_order) + 1
        node.order = Math.Max((node.left).order,
                              (node.right).order) + 1;
        // make pair of assigned value
        // and tree value
        v.Add(new Pair(node.order, node.data));
    }
}
// Used for sorting in ascending order
// of roll number
public static int compare(Pair a, Pair b)
{
    if(a.first != b.first)
        return (a.first - b.first);
    else
        return (a.second - b.second);
}
// Function to print leaf nodes in
// the given order
static void printLeafNodes(int n)
{
    // Sort the List in increasing order
    // of assigned node values
    v.Sort(compare);
    for (int i = 0; i     {
        if (i != v.Count - 1 &&
            v[i].first == v[i + 1].first)
            Console.Write(v[i].second + " ");
        else
            Console.Write(v[i].second + "\n");
    }
}
// Driver Code
public static void Main(String[] args)
{
    Node root = newNode(8, 0);
    root.left = newNode(3, 0);
    root.right = newNode(10, 0);
    root.left.left = newNode(1, 0);
    root.left.right = newNode(6, 0);
    root.right.left = newNode(14, 0);
    root.right.right = newNode(4, 0);
    root.left.left.left = newNode(7, 0);
    root.left.left.right = newNode(13, 0);
    int n = 9;
    Postorder(root);
    printLeafNodes(n);
}
}
// This code is contributed
// by Arnab Kundu


java 描述语言

Output: 

4 6 7 13 14
1 10
3
8


时间复杂度 : O(nlogn)
辅助空间 : O(n)，其中 n 是给定二叉树中的节点数。