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.
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分析:
由於是二叉搜索樹,那麼就滿足完全二叉樹的結構。只有可能最後一層的右邊會少幾個結點。
因此,如果是滿二叉樹的話,只需要求出整顆二叉搜索樹的層數h,然後利用滿二叉樹的性質:節點數爲2^h-1。
否則就只有分別求出左子樹的結點數與右子樹的結點數之和了。求這兩個字數的結點數的方法和判斷根節點一樣,也是先判斷是否是滿二叉樹,即遞歸調用即可。
以下是C++實習代碼:
/**//////////////////////84ms*/
/**
* 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 lef_height(TreeNode* root){ //求出左子樹的最大層次
int cnt = 0;
while(root != NULL){
cnt++;
root = root->left;
}
return cnt;
}
int rig_height(TreeNode* root){ //求出右子樹的最大層次
int cnt = 0;
while(root != NULL){
cnt++;
root = root->right;
}
return cnt;
}
int countNodes(TreeNode* root) {
int cnt = 0;
if(root == NULL)
return cnt;
int lef = lef_height(root);
int rig = rig_height(root);
if(lef == rig) //左右子樹的層次是相等的,說明是滿二叉樹,直接用公式計算節點總數
return (1<<lef)-1;
else //不是滿二叉數,那麼遞歸調用計算出左右字數的結點數再加上根節點數即可
return countNodes(root->left) + countNodes(root->right) + 1;
}
};