leetcode--Count Complete Tree Nodes
2017-08-08 14:53
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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.
题意:给定一棵完全二叉树,计算节点数目。
分类:二叉树
解法1:完全二叉树的定义是,除最后一层外,每一层上的节点数均达到最大值;在最后一层上只缺少右边的若干结点。
完全二叉树的一个特殊例子是满二叉树。
如果一棵树是满二叉树,那么我们可以用公式2^k-1计算它的节点数目。
如果不是满二叉树,我们要分别计算左子树和右子树,这是一个递归过程,最好左右子树和+1,为结果
根据上面的说法,我们可以先通过查找最左边节点的个数,最右边节点的个数
如果这两个数目相同,就是满二叉树,可以通过公式直接返回
如果不同,则递归分别计算左右子树的数目
[java] view
plain copy
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int countNodes(TreeNode root) {
if(root==null) return 0;
int left = getLeft(root.left)+1;
int right = getRight(root.right)+1;
if(left==right){//如果左右相等,就是满二叉树
return (2<<(left-1))-1;
}else{//如果左右不等,分别递归计算
return countNodes(root.left)+countNodes(root.right)+1;
}
}
/**
* 获得最左边节点的数目
*/
int getLeft(TreeNode root){
int left = 0;
while(root!=null){
left++;
root = root.left;
}
return left;
}
/**
* 获得最右边节点的数目
int getRight(TreeNode root){
int right = 0;
while(root!=null){
right++;
root = root.right;
}
return right;
}
}
原文链接http://blog.csdn.net/crazy__chen/article/details/47189559
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.
题意:给定一棵完全二叉树,计算节点数目。
分类:二叉树
解法1:完全二叉树的定义是,除最后一层外,每一层上的节点数均达到最大值;在最后一层上只缺少右边的若干结点。
完全二叉树的一个特殊例子是满二叉树。
如果一棵树是满二叉树,那么我们可以用公式2^k-1计算它的节点数目。
如果不是满二叉树,我们要分别计算左子树和右子树,这是一个递归过程,最好左右子树和+1,为结果
根据上面的说法,我们可以先通过查找最左边节点的个数,最右边节点的个数
如果这两个数目相同,就是满二叉树,可以通过公式直接返回
如果不同,则递归分别计算左右子树的数目
[java] view
plain copy
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int countNodes(TreeNode root) {
if(root==null) return 0;
int left = getLeft(root.left)+1;
int right = getRight(root.right)+1;
if(left==right){//如果左右相等,就是满二叉树
return (2<<(left-1))-1;
}else{//如果左右不等,分别递归计算
return countNodes(root.left)+countNodes(root.right)+1;
}
}
/**
* 获得最左边节点的数目
*/
int getLeft(TreeNode root){
int left = 0;
while(root!=null){
left++;
root = root.left;
}
return left;
}
/**
* 获得最右边节点的数目
int getRight(TreeNode root){
int right = 0;
while(root!=null){
right++;
root = root.right;
}
return right;
}
}
原文链接http://blog.csdn.net/crazy__chen/article/details/47189559
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