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Count Complete Tree Nodes.c
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Count Complete Tree Nodes.c
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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
Method One:
int leftlen(struct TreeNode* root){
struct TreeNode* temp;
int len=0;
temp=root;
while(temp!=NULL){
len++;
temp=temp->left;
}
return len;
}
int rightlen(struct TreeNode* root){
struct TreeNode* temp;
int len=0;
temp=root;
while(temp!=NULL){
len++;
temp=temp->right;
}
return len;
}
int isLeftRight(struct TreeNode* root, int fool, int hight){
struct TreeNode* temp;
temp=root;
while(temp!=NULL){
if(temp->right==NULL && fool<hight-1) return 1;
else if(temp->left==NULL && temp->right==NULL) return 0;
temp=temp->right;
fool++;
}
}
int countNodes(struct TreeNode* root) {
struct TreeNode* temp;
int sum,left,right,i,t,r[32],mark=0,fool,hight;
left=leftlen(root);
right=rightlen(root);
sum=(int )pow(2,(double) left)-1;
if(left==right) return sum;
hight=left;
temp=root;
fool=0;
while(temp!=NULL){
if(temp->left!=NULL && temp->right!=NULL){
t=isLeftRight(temp->left,fool+1,hight);
}else if(temp->left!=NULL && temp->right==NULL){
r[fool]=1;
fool++;
break;
}else if(temp->left==NULL && temp->right==NULL){
mark=1;
r[fool]=0;
fool++;
break;
}
if(t==1){
temp=temp->left;
r[fool]=1;
}else{
temp=temp->right;
r[fool]=0;
}
fool++;
}
t=0;
for(i=0;i<fool;i++){
t=t*2+r[i];
}
if(mark==1) t+=2;
return sum-t;
}
Method Two:
int leftlen(struct TreeNode* root){
struct TreeNode* temp;
int len=0;
temp=root;
while(temp!=NULL){
len++;
temp=temp->left;
}
return len;
}
int rightlen(struct TreeNode* root){
struct TreeNode* temp;
int len=0;
temp=root;
while(temp!=NULL){
len++;
temp=temp->right;
}
return len;
}
int countNodes(struct TreeNode* root) {
struct TreeNode* temp;
int sum,left,right;
left=leftlen(root);
right=rightlen(root);
if(left==right) return (1<<left)-1;
else return 1+ countNodes(root->left)+countNodes(root->right);
}