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完全二叉树

在完全二叉树中,除了最底层节点可能没填满外,其余每层节点数都达到最大值,并且最下面一层的节点都集中在该层最左边的若干位置。若最底层为第 h 层,则该层包含 1~ 2h 个节点。

完全二叉树的节点个数

给你一棵 完全二叉树 的根节点 root ,求出该树的节点个数。

  • 输入:root = [1,2,3,4,5,6]

    graph TB;
        A((1))-->B((2))
        A-->C((3));
        B-->E((4))
        B-->F((5))
        C-->H((6))

    输出:6

  • 输入:root = []

    输出:0

  • 输入:root = [1]

    输出:1

二叉树节点定义:

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struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode() : val(0), left(nullptr), right(nullptr) {}
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

遍历

时间复杂度 \(O(n)\),空间复杂度 \(O(\log\ n)\)

递归

时间复杂度 \(O(\log^2\ n)\),空间复杂度 \(O(\log\ n)\)

int countNodes(TreeNode* root) {
    return countNodesRecur(root);
}

int countNodesRecur(TreeNode* root) {
    if (!root) return 0;

    int leftDepth = getDepth(root, 0);
    int rightDepth = getDepth(root, 1);

    if (leftDepth == rightDepth)
    {
        return (1 << leftDepth) - 1;
    }

    return countNodesRecur(root->left) + countNodesRecur(root->right) + 1;
}

int getDepth(TreeNode* root, int right) {
    int depth = 0;
    while (root)
    {
        depth++;
        root = right ? root->right : root->left;
    }
    return depth;
}

二分查找

时间复杂度 \(O(\log^2\ n)\),空间复杂度 \(O(1)\)

int countNodes(TreeNode* root) {
    return countNodesBinarySearch(root);
}

int countNodesBinarySearch(TreeNode* root) {
    int h = getDepth(root, 0) - 1;
    if (h < 0) return 0;

    int low = 0;
    int high = (1 << h) - 1;

    // [low, high]
    while (low < high)
    {
        int mid = (high - low + 1) / 2 + low;
        TreeNode* node = root;

        for (int i = h - 1;i >= 0;i--)
        {
            int cmd = (mid >> i) & 1;
            node = cmd ? node->right : node->left;
        }

        if (node) low = mid;
        else high = mid - 1;
    }

    return (low + 1) + ((1 << h) - 1);
}

int getDepth(TreeNode* root, int right) {
    int depth = 0;
    while (root)
    {
        depth++;
        root = right ? root->right : root->left;
    }
    return depth;
}