给定一棵二叉树。首先打印所有叶节点，然后从树中删除所有叶节点，然后打印所有新形成的叶节点并继续执行此操作，直到从树中删除所有节点为止。
例子：
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
方法：想法是执行简单的dfs并根据以下条件为每个节点分配不同的值：
- 最初将所有节点分配为0。
- 现在，为所有节点分配值为(两个子项的最大值)+1 。
DFS之前的树：将临时值零分配给所有节点。 
DFS之后的树：节点被分配为(两个子节点的最大值)+1的值。 
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之后，该任务现在减少了，以分配给它们的节点值的升序打印树。
下面是上述方法的实现：
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;
};
// Utiltiy 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 < n; i++) {
if (v[i].first == v[i + 1].first)
cout << v[i].second << " ";
else
cout << v[i].second << "\n";
}
}
// 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
// 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;
}
}
// Utiltiy 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 < v.size(); 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 Python3
# 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 SHUBHAMSINGH10C#
// 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;
}
}
// Utiltiy 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 < v.Count; 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 输出：
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是给定二叉树中节点的数量。