来自http://www.geeksforgeeks.org/find-a-pair-with-given-sum-in-bst/
/* In a balanced binary search tree isPairPresent two element which sums to
a given value time O(n) space O(logn) */
#include <stdio.h>
#include <stdlib.h>
#define MAX_SIZE 100
// A BST node
struct node
{
int val;
struct node *left, *right;
};
// Stack type
struct Stack
{
int size;
int top;
struct node* *array;
};
// A utility function to create a stack of given size
struct Stack* createStack(int size)
{
struct Stack* stack =
(struct Stack*) malloc(sizeof(struct Stack));
stack->size = size;
stack->top = -1;
stack->array =
(struct node**) malloc(stack->size * sizeof(struct node*));
return stack;
}
// BASIC OPERATIONS OF STACK
int isFull(struct Stack* stack)
{ return stack->top - 1 == stack->size; }
int isEmpty(struct Stack* stack)
{ return stack->top == -1; }
void push(struct Stack* stack, struct node* node)
{
if (isFull(stack))
return;
stack->array[++stack->top] = node;
}
struct node* pop(struct Stack* stack)
{
if (isEmpty(stack))
return NULL;
return stack->array[stack->top--];
}
// Returns true if a pair with target sum exists in BST, otherwise false
bool isPairPresent(struct node *root, int target)
{
// Create two stacks. s1 is used for normal inorder traversal
// and s2 is used for reverse inorder traversal
struct Stack* s1 = createStack(MAX_SIZE);
struct Stack* s2 = createStack(MAX_SIZE);
// Note the sizes of stacks is MAX_SIZE, we can find the tree size and
// fix stack size as O(Logn) for balanced trees like AVL and Red Black
// tree. We have used MAX_SIZE to keep the code simple
// done1, val1 and curr1 are used for normal inorder traversal using s1
// done2, val2 and curr2 are used for reverse inorder traversal using s2
bool done1 = false, done2 = false;
int val1 = 0, val2 = 0;
struct node *curr1 = root, *curr2 = root;
// The loop will break when we either find a pair or one of the two
// traversals is complete
while (1)
{
// Find next node in normal Inorder traversal. See following post
// http://www.geeksforgeeks.org/inorder-tree-traversal-without-recursion/
while (done1 == false)
{
if (curr1 != NULL)
{
push(s1, curr1);
curr1 = curr1->left;
}
else
{
if (isEmpty(s1))
done1 = 1;
else
{
curr1 = pop(s1);
val1 = curr1->val;
curr1 = curr1->right;
done1 = 1;
}
}
}
// Find next node in REVERSE Inorder traversal. The only
// difference between above and below loop is, in below loop
// right subtree is traversed before left subtree
while (done2 == false)
{
if (curr2 != NULL)
{
push(s2, curr2);
curr2 = curr2->right;
}
else
{
if (isEmpty(s2))
done2 = 1;
else
{
curr2 = pop(s2);
val2 = curr2->val;
curr2 = curr2->left;
done2 = 1;
}
}
}
// If we find a pair, then print the pair and return. The first
// condition makes sure that two same values are not added
if ((val1 != val2) && (val1 + val2) == target)
{
printf("\n Pair Found: %d + %d = %d\n", val1, val2, target);
return true;
}
// If sum of current values is smaller, then move to next node in
// normal inorder traversal
else if ((val1 + val2) < target)
done1 = false;
// If sum of current values is greater, then move to next node in
// reverse inorder traversal
else if ((val1 + val2) > target)
done2 = false;
// If any of the inorder traversals is over, then there is no pair
// so return false
if (val1 >= val2)
return false;
}
}
// A utility function to create BST node
struct node * NewNode(int val)
{
struct node *tmp = (struct node *)malloc(sizeof(struct node));
tmp->val = val;
tmp->right = tmp->left =NULL;
return tmp;
}
// Driver program to test above functions
int main()
{
/*
15
/ \
10 20
/ \ / \
8 12 16 25 */
struct node *root = NewNode(15);
root->left = NewNode(10);
root->right = NewNode(20);
root->left->left = NewNode(8);
root->left->right = NewNode(12);
root->right->left = NewNode(16);
root->right->right = NewNode(25);
int target = 28;
if (isPairPresent(root, target) == false)
printf("\n No such values are found\n");
getchar();
return 0;
}