检查两棵二叉树的叶子遍历是否相同？
叶遍历是从左到右遍历的叶序列。问题是检查两个给定二叉树的叶子遍历是否相同。
预期时间复杂度 O(n)。预期的辅助空间 O(h1 + h2)，其中 h1 和 h2 是两棵二叉树的高度。
例子：
Input: Roots of below Binary Trees
1
/ \
2 3
/ / \
4 6 7
0
/ \
5 8
\ / \
4 6 7
Output: same
Leaf order traversal of both trees is 4 6 7
Input: Roots of below Binary Trees
0
/ \
1 2
/ \
8 9
1
/ \
4 3
\ / \
8 2 9
Output: Not Same
Leaf traversals of two trees are different.
For first, it is 8 9 2 and for second it is
8 2 9我们强烈建议您最小化您的浏览器并首先自己尝试。
一个简单的解决方案是遍历第一棵树并将叶子从左到右存储在一个数组中。然后遍历其他树并将叶子存储在另一个数组中。最后比较两个数组。如果两个数组相同，则返回 true。
上述解决方案需要 O(m+n) 额外空间，其中 m 和 n 分别是第一棵树和第二棵树中的节点。
如何检查 O(h1 + h2) 空间？
这个想法是使用迭代遍历。同时遍历两棵树，在两棵树中寻找叶子节点并比较找到的叶子。所有叶子必须匹配。
算法：
1. Create empty stacks stack1 and stack2
for iterative traversals of tree1 and tree2
2. insert (root of tree1) in stack1
insert (root of tree2) in stack2
3. Stores current leaf nodes of tree1 and tree2
temp1 = (root of tree1)
temp2 = (root of tree2)
4. Traverse both trees using stacks
while (stack1 and stack2 parent empty)
{
// Means excess leaves in one tree
if (if one of the stacks are empty)
return false
// get next leaf node in tree1
temp1 = stack1.pop()
while (temp1 is not leaf node)
{
push right child to stack1
push left child to stack1
}
// get next leaf node in tree2
temp2 = stack2.pop()
while (temp2 is not leaf node)
{
push right child to stack2
push left child to stack2
}
// If leaves do not match return false
if (temp1 != temp2)
return false
}
5. If all leaves matched, return true下面是上述算法的Java实现。
C++
// C++ code to check if leaf traversals
// of two Binary Trees are same or not.
#include
using namespace std;
// Binary Tree Node
struct Node {
int data;
Node* left;
Node* right;
};
// Returns new Node with data as
// input to below function.
Node* newNode(int d)
{
Node* temp = new Node;
temp->data = d;
temp->left = NULL;
temp->right = NULL;
return temp;
}
// checks if a given node is leaf or not.
bool isLeaf(Node* root)
{
if (root == NULL)
return false;
if (!root->left && !root->right)
return true;
return false;
}
// iterative function.
// returns true if leaf traversals
// are same, else false.
bool isSame(Node* root1, Node* root2)
{
stack s1;
stack s2;
// push root1 to empty stack s1.
s1.push(root1);
// push root2 to empty stack s2.
s2.push(root2);
// loop until either of stacks are non-empty.
while (!s1.empty() || !s2.empty())
{
// this means one of the stacks has
// extra leaves, hence return false.
if (s1.empty() || s2.empty())
return false;
Node* temp1 = s1.top();
s1.pop();
while (temp1 != NULL && !isLeaf(temp1))
{
// Push right child if exists
if (temp1->right)
s1.push(temp1->right);
// Push left child if exists
if (temp1->left)
s1.push(temp1->left);
// Note that right child(if exists)
// is pushed before left child(if exists).
temp1 = s1.top();
s1.pop();
}
Node* temp2 = s2.top();
s2.pop();
while (temp2 != NULL && !isLeaf(temp2))
{
// Push right child if exists
if (temp2->right)
s2.push(temp2->right);
// Push left child if exists
if (temp2->left)
s2.push(temp2->left);
temp2 = s2.top();
s2.pop();
}
if (!temp1 && temp2)
return false;
if (temp1 && !temp2)
return false;
if (temp1 && temp2) {
return temp1->data == temp2->data;
}
}
// all leaves are matched
return true;
}
// Driver Code
int main()
{
Node* root1 = newNode(1);
root1->left = newNode(2);
root1->right = newNode(3);
root1->left->left = newNode(4);
root1->right->left = newNode(6);
root1->right->right = newNode(7);
Node* root2 = newNode(0);
root2->left = newNode(1);
root2->right = newNode(5);
root2->left->right = newNode(4);
root2->right->left = newNode(6);
root2->right->right = newNode(7);
if (isSame(root1, root2))
cout << "Same";
else
cout << "Not Same";
return 0;
}
// This code is contributed
// by AASTHA VARMA Java
// Java program to check if two Leaf Traversal of
// Two Binary Trees is same or not
import java.util.*;
import java.lang.*;
import java.io.*;
// Binary Tree node
class Node
{
int data;
Node left, right;
public Node(int x)
{
data = x;
left = right = null;
}
public boolean isLeaf()
{
return (left == null && right == null);
}
}
class LeafOrderTraversal
{
// Returns true of leaf traversal of two trees is
// same, else false
public static boolean isSame(Node root1, Node root2)
{
// Create empty stacks. These stacks are going
// to be used for iterative traversals.
Stack s1 = new Stack();
Stack s2 = new Stack();
s1.push(root1);
s2.push(root2);
// Loop until either of two stacks is not empty
while (!s1.empty() || !s2.empty())
{
// If one of the stacks is empty means other
// stack has extra leaves so return false
if (s1.empty() || s2.empty())
return false;
Node temp1 = s1.pop();
while (temp1 != null && !temp1.isLeaf())
{
// Push right and left children of temp1.
// Note that right child is inserted
// before left
if (temp1.right != null)
s1.push(temp1.right);
if (temp1.left != null)
s1.push(temp1.left);
temp1 = s1.pop();
}
// same for tree2
Node temp2 = s2.pop();
while (temp2 != null && !temp2.isLeaf())
{
if (temp2.right != null)
s2.push(temp2.right);
if (temp2.left != null)
s2.push(temp2.left);
temp2 = s2.pop();
}
// If one is null and other is not, then
// return false
if (temp1 == null && temp2 != null)
return false;
if (temp1 != null && temp2 == null)
return false;
// If both are not null and data is not
// same return false
if (temp1 != null && temp2 != null)
{
if (temp1.data != temp2.data)
return false;
}
}
// If control reaches this point, all leaves
// are matched
return true;
}
// Driver code
public static void main(String[] args)
{
// Let us create trees in above example 1
Node root1 = new Node(1);
root1.left = new Node(2);
root1.right = new Node(3);
root1.left.left = new Node(4);
root1.right.left = new Node(6);
root1.right.right = new Node(7);
Node root2 = new Node(0);
root2.left = new Node(1);
root2.right = new Node(5);
root2.left.right = new Node(4);
root2.right.left = new Node(6);
root2.right.right = new Node(7);
if (isSame(root1, root2))
System.out.println("Same");
else
System.out.println("Not Same");
}
} Python3
# Python3 program to check if two Leaf
# Traversal of Two Binary Trees is same or not
# Binary Tree node
class Node:
def __init__(self, x):
self.data = x
self.left = self.right = None
def isLeaf(self):
return (self.left == None and
self.right == None)
# Returns true of leaf traversal of
# two trees is same, else false
def isSame(root1, root2):
# Create empty stacks. These stacks are going
# to be used for iterative traversals.
s1 = []
s2 = []
s1.append(root1)
s2.append(root2)
# Loop until either of two stacks
# is not empty
while (len(s1) != 0 or len(s2) != 0):
# If one of the stacks is empty means other
# stack has extra leaves so return false
if (len(s1) == 0 or len(s2) == 0):
return False
temp1 = s1.pop(-1)
while (temp1 != None and not temp1.isLeaf()):
# append right and left children of temp1.
# Note that right child is inserted
# before left
if (temp1.right != None):
s1.append(temp1. right)
if (temp1.left != None):
s1.append(temp1.left)
temp1 = s1.pop(-1)
# same for tree2
temp2 = s2.pop(-1)
while (temp2 != None and not temp2.isLeaf()):
if (temp2.right != None):
s2.append(temp2.right)
if (temp2.left != None):
s2.append(temp2.left)
temp2 = s2.pop(-1)
# If one is None and other is not,
# then return false
if (temp1 == None and temp2 != None):
return False
if (temp1 != None and temp2 == None):
return False
# If both are not None and data is
# not same return false
if (temp1 != None and temp2 != None):
if (temp1.data != temp2.data):
return False
# If control reaches this point,
# all leaves are matched
return True
# Driver Code
if __name__ == '__main__':
# Let us create trees in above example 1
root1 = Node(1)
root1.left = Node(2)
root1.right = Node(3)
root1.left.left = Node(4)
root1.right.left = Node(6)
root1.right.right = Node(7)
root2 = Node(0)
root2.left = Node(1)
root2.right = Node(5)
root2.left.right = Node(4)
root2.right.left = Node(6)
root2.right.right = Node(7)
if (isSame(root1, root2)):
print("Same")
else:
print("Not Same")
# This code is contributed by pranchalKC#
// C# program to check if two Leaf Traversal
// of Two Binary Trees is same or not
using System;
using System.Collections.Generic;
// Binary Tree node
public class Node {
public int data;
public Node left, right;
public Node(int x)
{
data = x;
left = right = null;
}
public virtual bool Leaf
{
get {
return (left == null && right == null);
}
}
}
class GFG {
// Returns true of leaf traversal of
// two trees is same, else false
public static bool isSame(Node root1, Node root2)
{
// Create empty stacks. These stacks
// are going to be used for iterative
// traversals.
Stack s1 = new Stack();
Stack s2 = new Stack();
s1.Push(root1);
s2.Push(root2);
// Loop until either of two stacks
// is not empty
while (s1.Count > 0 || s2.Count > 0)
{
// If one of the stacks is empty means other
// stack has extra leaves so return false
if (s1.Count == 0 || s2.Count == 0)
{
return false;
}
Node temp1 = s1.Pop();
while (temp1 != null && !temp1.Leaf)
{
// Push right and left children of temp1.
// Note that right child is inserted
// before left
if (temp1.right != null)
{
s1.Push(temp1.right);
}
if (temp1.left != null)
{
s1.Push(temp1.left);
}
temp1 = s1.Pop();
}
// same for tree2
Node temp2 = s2.Pop();
while (temp2 != null && !temp2.Leaf)
{
if (temp2.right != null)
{
s2.Push(temp2.right);
}
if (temp2.left != null)
{
s2.Push(temp2.left);
}
temp2 = s2.Pop();
}
// If one is null and other is not,
// then return false
if (temp1 == null && temp2 != null)
{
return false;
}
if (temp1 != null && temp2 == null)
{
return false;
}
// If both are not null and data
// is not same return false
if (temp1 != null && temp2 != null)
{
if (temp1.data != temp2.data)
{
return false;
}
}
}
// If control reaches this point,
// all leaves are matched
return true;
}
// Driver Code
public static void Main(string[] args)
{
// Let us create trees in above example 1
Node root1 = new Node(1);
root1.left = new Node(2);
root1.right = new Node(3);
root1.left.left = new Node(4);
root1.right.left = new Node(6);
root1.right.right = new Node(7);
Node root2 = new Node(0);
root2.left = new Node(1);
root2.right = new Node(5);
root2.left.right = new Node(4);
root2.right.left = new Node(6);
root2.right.right = new Node(7);
if (isSame(root1, root2)) {
Console.WriteLine("Same");
}
else {
Console.WriteLine("Not Same");
}
}
}
// This code is contributed by Shrikant13 Javascript
输出
Same