222. Count Complete Tree Nodes -- 求完全二叉树节点个数

Given a complete binary tree, count the number of nodes.

Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.

1. 递归

//return -1 if it is not.
int isCompleteTree(TreeNode* root) {
if (!root) return 0;

int cnt = 1;
TreeNode *left = root, *right = root;
for(; left && right; left=left->left, right=right->right) {
cnt *= 2;
}

if (left!=NULL || right!=NULL) {
return -1;
}
return cnt-1;
}

int countNodes(TreeNode* root) {
int cnt = isCompleteTree(root);
if (cnt != -1) return cnt;
int leftCnt = countNodes(root->left);
int rightCnt = countNodes(root->right);
return leftCnt + rightCnt + 1;
}

2. 非递归

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int countNodes(TreeNode* root) {
int ans = 0, lh = 0, rh = 0;
TreeNode *p = root;
while(p)
{
//先求p的左右子树的最大深度。若相等，则p加左子树节点个数为1<<lh，再求右子树；若不等，则p加右子树节点个数为1<<rh，再求左子树。
if(!lh) //若lh不为0，说明lh由之前的lh--得到，是已知的。
{
for(TreeNode *t = p->left; t; t = t->left)
lh++;
}
for(TreeNode *t = p->right; t; t = t->left)
rh++;
if(lh == rh)
{
ans += 1 << lh;
p = p->right;
}
else
{
ans += 1 << rh;
p = p->left;
}
lh--;
rh = 0;
}
return ans;
}
};