[LeetCode] 222. Count Complete Tree Nodes
阿新 • • 發佈:2018-11-09
Count Complete Tree Nodes
Given a complete binary tree, count the number of nodes.
Note:
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.
Example:
Input:
1
/ \
2 3
/ \ /
4 5 6
Output: 6
解析
求完整二叉樹的節點個數,直接遞迴左右節點個數。分別找出以當前節點為根節點的左子樹和右子樹的高度並對比,如果相等,則說明是滿二叉樹,直接返回節點個數,如果不相等,則節點個數為左子樹的節點個數加上右子樹的節點個數再加1(根節點),其中左右子樹節點個數的計算可以使用遞迴來計算。
程式碼
class Solution {
public:
int countNodes(TreeNode* root) {
if(!root) return 0;
TreeNode* leftnode = root->left ;
TreeNode* rightnode = root->right;
int lefth,righth;
lefth = righth = 0;
while(leftnode){
lefth ++;
leftnode = leftnode->left;
}
while(rightnode){
righth ++;
rightnode = rightnode->right;
}
if (lefth == righth) return pow(2, lefth+1) -1;
else return countNodes(root->left) + countNodes(root->right) +1;
}
};