二叉树的倾斜度
给定一棵二叉树，返回整棵树的倾斜度。树节点的倾斜度定义为所有左子树节点值之和与所有右子树节点值之和之间的绝对差。空节点被分配倾斜为零。因此，整棵树的倾斜度定义为所有节点的倾斜度之和。
例子：
Input :
1
/ \
2 3
Output : 1
Explanation:
Tilt of node 2 : 0
Tilt of node 3 : 0
Tilt of node 1 : |2-3| = 1
Tilt of binary tree : 0 + 0 + 1 = 1
Input :
4
/ \
2 9
/ \ \
3 5 7
Output : 15
Explanation:
Tilt of node 3 : 0
Tilt of node 5 : 0
Tilt of node 7 : 0
Tilt of node 2 : |3-5| = 2
Tilt of node 9 : |0-7| = 7
Tilt of node 4 : |(3+5+2)-(9+7)| = 6
Tilt of binary tree : 0 + 0 + 0 + 2 + 7 + 6 = 15这个想法是递归遍历树。遍历时，我们跟踪两件事，以当前节点为根的子树的总和，当前节点的倾斜度。需要总和来计算父级的倾斜度。
C++
// CPP Program to find Tilt of Binary Tree
#include
using namespace std;
/* A binary tree node has data, pointer to
left child and a pointer to right child */
struct Node {
int val;
struct Node *left, *right;
};
/* Recursive function to calculate Tilt of
whole tree */
int traverse(Node* root, int* tilt)
{
if (!root)
return 0;
// Compute tilts of left and right subtrees
// and find sums of left and right subtrees
int left = traverse(root->left, tilt);
int right = traverse(root->right, tilt);
// Add current tilt to overall
*tilt += abs(left - right);
// Returns sum of nodes under current tree
return left + right + root->val;
}
/* Driver function to print Tilt of whole tree */
int Tilt(Node* root)
{
int tilt = 0;
traverse(root, &tilt);
return tilt;
}
/* Helper function that allocates a
new node with the given data and
NULL left and right pointers. */
Node* newNode(int data)
{
Node* temp = new Node;
temp->val = data;
temp->left = temp->right = NULL;
return temp;
}
// Driver code
int main()
{
/* Let us construct a Binary Tree
4
/ \
2 9
/ \ \
3 5 7 */
Node* root = NULL;
root = newNode(4);
root->left = newNode(2);
root->right = newNode(9);
root->left->left = newNode(3);
root->left->right = newNode(8);
root->right->right = newNode(7);
cout << "The Tilt of whole tree is " << Tilt(root);
return 0;
} Java
// Java Program to find Tilt of Binary Tree
import java.util.*;
class GfG {
/* A binary tree node has data, pointer to
left child and a pointer to right child */
static class Node {
int val;
Node left, right;
}
/* Recursive function to calculate Tilt of
whole tree */
static class T{
int tilt = 0;
}
static int traverse(Node root, T t )
{
if (root == null)
return 0;
// Compute tilts of left and right subtrees
// and find sums of left and right subtrees
int left = traverse(root.left, t);
int right = traverse(root.right, t);
// Add current tilt to overall
t.tilt += Math.abs(left - right);
// Returns sum of nodes under current tree
return left + right + root.val;
}
/* Driver function to print Tilt of whole tree */
static int Tilt(Node root)
{
T t = new T();
traverse(root, t);
return t.tilt;
}
/* Helper function that allocates a
new node with the given data and
NULL left and right pointers. */
static Node newNode(int data)
{
Node temp = new Node();
temp.val = data;
temp.left = null;
temp.right = null;
return temp;
}
// Driver code
public static void main(String[] args)
{
/* Let us construct a Binary Tree
4
/ \
2 9
/ \ \
3 5 7 */
Node root = null;
root = newNode(4);
root.left = newNode(2);
root.right = newNode(9);
root.left.left = newNode(3);
root.left.right = newNode(8);
root.right.right = newNode(7);
System.out.println("The Tilt of whole tree is " + Tilt(root));
}
}Python3
# Python3 Program to find Tilt of
# Binary Tree
# class that allocates a new node
# with the given data and
# None left and right pointers.
class newNode:
def __init__(self, data):
self.val = data
self.left = self.right = None
# Recursive function to calculate
# Tilt of whole tree
def traverse(root, tilt):
if (not root):
return 0
# Compute tilts of left and right subtrees
# and find sums of left and right subtrees
left = traverse(root.left, tilt)
right = traverse(root.right, tilt)
# Add current tilt to overall
tilt[0] += abs(left - right)
# Returns sum of nodes under
# current tree
return left + right + root.val
# Driver function to print Tilt
# of whole tree
def Tilt(root):
tilt = [0]
traverse(root, tilt)
return tilt[0]
# Driver code
if __name__ == '__main__':
# Let us construct a Binary Tree
# 4
# / \
# 2 9
# / \ \
# 3 5 7
root = None
root = newNode(4)
root.left = newNode(2)
root.right = newNode(9)
root.left.left = newNode(3)
root.left.right = newNode(8)
root.right.right = newNode(7)
print("The Tilt of whole tree is",
Tilt(root))
# This code is contributed by PranchalKC#
// C# Program to find Tilt of Binary Tree
using System;
class GfG
{
/* A binary tree node has data, pointer to
left child and a pointer to right child */
public class Node
{
public int val;
public Node left, right;
}
/* Recursive function to calculate Tilt of
whole tree */
public class T
{
public int tilt = 0;
}
static int traverse(Node root, T t )
{
if (root == null)
return 0;
// Compute tilts of left and right subtrees
// and find sums of left and right subtrees
int left = traverse(root.left, t);
int right = traverse(root.right, t);
// Add current tilt to overall
t.tilt += Math.Abs(left - right);
// Returns sum of nodes under current tree
return left + right + root.val;
}
/* Driver function to print Tilt of whole tree */
static int Tilt(Node root)
{
T t = new T();
traverse(root, t);
return t.tilt;
}
/* Helper function that allocates a
new node with the given data and
NULL left and right pointers. */
static Node newNode(int data)
{
Node temp = new Node();
temp.val = data;
temp.left = null;
temp.right = null;
return temp;
}
// Driver code
public static void Main(String[] args)
{
/* Let us construct a Binary Tree
4
/ \
2 9
/ \ \
3 5 7 */
Node root = null;
root = newNode(4);
root.left = newNode(2);
root.right = newNode(9);
root.left.left = newNode(3);
root.left.right = newNode(8);
root.right.right = newNode(7);
Console.WriteLine("The Tilt of whole tree is " + Tilt(root));
}
}
// This code contributed by Rajput-JiJavascript
输出：
The Tilt of whole tree is 15复杂性分析：
- 时间复杂度： O(n)，其中 n 是二叉树的节点数。
- 辅助空间： O(n)，在最坏的情况下，二叉树的深度为 n。