Friday, November 27, 2015

[itint5]完全二叉树节点个数的统计 - 阿牧遥 - 博客园


[itint5]完全二叉树节点个数的统计 - 阿牧遥 - 博客园
这题是利用完全二叉树的性质计算节点数目。那么是通过比较左右子树的最左结点的高度来看那边是满的,然后递归计算。

 给定一棵完全二叉树的根结点,统计该树的结点总数。
 树结点类型名为TreeNode,您没必要知道该类型的定义,请使用下面的方法得到某个结点的左,右儿子结点,以及判断某结点是否为NULL。
https://github.com/AnnieKim/ITint5/blob/master/004_%E7%BB%9F%E8%AE%A1%E5%AE%8C%E5%85%A8%E4%BA%8C%E5%8F%89%E6%A0%91%E7%BB%93%E7%82%B9%E6%95%B0.cpp
 //获得结点node的左儿子结点
 TreeNode getLeftChildNode(TreeNode node);
 //获得结点node的右儿子结点
 TreeNode getRightChildNode(TreeNode node);
 //判断结点node是否为NULL
 int isNullNode(TreeNode node);
 Solution: 这里要充分利用完全二叉树的性质。
           根据左右子树的高度,可以判断出最后一层最右叶节点的位置,就可以进行二分。
           复杂度O(lgn^2)。
*/

// 树的高度,沿着左孩子
int getHeight(TreeNode node)
{
    int height = 0;
    while (!isNullNode(node)) {
        height++;
        node = getLeftChildNode(node);
    }
    return height;
}

// 高度为height的满二叉树节点个数
int count_perfect_binary_tree_nodes(int height)
{
    return (int)pow(2, height) - 1;
}

int count_complete_binary_tree_nodes(TreeNode root)
{
    int count = 0;
    while (!isNullNode(root))
    {
        int left = getHeight(getLeftChildNode(root));
        int right = getHeight(getRightChildNode(root));
        if (left == right)
        {
            count += count_perfect_binary_tree_nodes(left) + 1;
            root = getRightChildNode(root);
        }
        else
        {
            count += count_perfect_binary_tree_nodes(right) + 1;
            root = getLeftChildNode(root);
        }
    }
    return count;
}
http://lixinzhang.github.io/itint5-mian-shi-ti-zong-jie.html
给定一棵完全二叉树(查看定义)的根结点,统计该树的结点总数。
树结点类型名为TreeNode,您没必要知道该类型的定义,请使用下面的方法得到某个结点的左,右儿子结点,以及判断某结点是否为NULL。

分析

O(n)的方法肯定是不行的,因为没有借助完全二叉树的特性。 于是,考虑计算最左边树高度,与最右边树高度,如果两个值相等,那么这棵子树的为一颗满二叉树,其节点个数自然就是已知的了,为2height1.
递归去做。传说中hulu的面试题。

//使用getLeftChildNode(TreeNode)获得左儿子结点
//使用getRightChildNode(TreeNode)获得右儿子结点 //使用isNullNode(TreeNode)判断结点是否为空 int count_complete_binary_tree_nodes(TreeNode root) { if(isNullNode(root)) return 0; TreeNode left = getLeftChildNode(root); int left_height = 0; while(!isNullNode(left)){ left = getLeftChildNode(left); left_height += 1; } int right_height = 0; TreeNode right = getRightChildNode(root); while(!isNullNode(right)){ right = getRightChildNode(right); right_height += 1; } if(left_height == right_height){ return (2 << left_height) -1; } return count_complete_binary_tree_nodes(getLeftChildNode(root)) + count_complete_binary_tree_nodes(getRightChildNode(root)) + 1; }

//使用getLeftChildNode(TreeNode)获得左儿子结点
//使用getRightChildNode(TreeNode)获得右儿子结点
//使用isNullNode(TreeNode)判断结点是否为空
int get_left_height(TreeNode root) {
    if (isNullNode(root)) {
        return 0;
    else {
        return get_left_height(getLeftChildNode(root)) + 1;
    }
}
int count_complete_binary_tree_nodes(TreeNode root) {
    if (isNullNode(root))
        return 0;
    TreeNode leftNode = getLeftChildNode(root);
    TreeNode rightNode = getRightChildNode(root);
    int lheight = get_left_height(leftNode);
    int rheight = get_left_height(rightNode);
    if (lheight == rheight) {
        return (1 << lheight) + count_complete_binary_tree_nodes(rightNode);
    else {
        return (1 << rheight) + count_complete_binary_tree_nodes(leftNode);
    }
}


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