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[LeetCode] 865. Smallest Subtree with all the Deepest Nodes #865

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grandyang opened this issue May 30, 2019 · 0 comments
Open

[LeetCode] 865. Smallest Subtree with all the Deepest Nodes #865

grandyang opened this issue May 30, 2019 · 0 comments

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@grandyang
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grandyang commented May 30, 2019

Given a binary tree rooted at root, the depth of each node is the shortest distance to the root.

A node is  deepest  if it has the largest depth possible among any node in the entire tree.

The subtree of a node is that node, plus the set of all descendants of that node.

Return the node with the largest depth such that it contains all the deepest nodes in its subtree.

Example 1:

Input: [3,5,1,6,2,0,8,null,null,7,4]
Output: [2,7,4]
Explanation:

We return the node with value 2, colored in yellow in the diagram.
The nodes colored in blue are the deepest nodes of the tree.
The input "[3, 5, 1, 6, 2, 0, 8, null, null, 7, 4]" is a serialization of the given tree.
The output "[2, 7, 4]" is a serialization of the subtree rooted at the node with value 2.
Both the input and output have TreeNode type.

Note:

  • The number of nodes in the tree will be between 1 and 500.
  • The values of each node are unique.

这道题给了我们一棵二叉树,让我们找包含所有最深结点的最小子树,就是返回这棵最小子树的根结点。题目中给了一个例子,因为有图,所以可以很直接的看出来最深的结点是7和4,那么包含这两个结点的最小子树的根结点是2,返回即可。其实最深的结点不一定只有两个,可能有很多个,比如对于一棵完全二叉树,即把例子图中的结点7和4去掉后,此时最深的结点就有四个,分别是6,2,0,8,都包含这些结点的子树就是原树本身了,要返回根结点。

通过上述分析,可以发现,子树的最大深度很重要,对于一棵完全二叉树来说,根结点的左右子树的最大深度一定是相同的,此时直接返回根结点即可。若左右子树的最大深度不同,则最深结点一定位于深度大的子树中,可以对其调用递归函数。所以只需要写一个计算最大深度的递归函数,来计算左右子树的最大深度差,再根据这个差值来决定对谁调用当前的递归函数,两个递归函数相互缠绕,画面美极了,参见代码如下:

解法一:

class Solution {
public:
    TreeNode* subtreeWithAllDeepest(TreeNode* root) {
        int diff = depth(root->left) - depth(root->right);
        return (diff == 0) ? root : subtreeWithAllDeepest(diff > 0 ? root->left : root->right);
    }
    int depth(TreeNode* node) {
        return !node ? 0 : max(depth(node->left), depth(node->right)) + 1;
    }
};

上面的解法其实并不高效,因为对于每个结点,都要统计其左右子树的最大深度,有大量的重复计算存在,我们来尝试提高时间复杂度,就不可避免的要牺牲一些空间。递归函数需要返回一个 pair,由每个结点的最大深度,以及包含最深结点的最小子树组成。所以在原函数中,对根结点调用递归函数,并取返回的 pair 中的第二项。

在递归函数中,首先判断结点是否存在,为空的话直接返回一个 {0, NULL} 对儿。否则分别对左右子结点调用递归函数,将各自的返回的 pair 存入 left 和 right 中,然后先分别在 left 和 right 中取出左右子树的最大深度 d1 和 d2,之后就要建立返回值的 pair,第一项为当前结点的最大深度,由左右子树中的最大深度加1组成,而包含最深结点的最小子树由 d1 和 d2 值的大小决定,若 d1>d2,则为 left.second,否则为 right.second,这样我们就把原本的两个递归,揉合到了一个递归函数中,大大提高了运行效率,参见代码如下:

解法二:

class Solution {
public:
    TreeNode* subtreeWithAllDeepest(TreeNode* root) {
        return helper(root).second;
    }
    pair<int, TreeNode*> helper(TreeNode* node) {
        if (!node) return {0, NULL};
        auto left = helper(node->left), right = helper(node->right);
        int d1 = left.first, d2 = right.first;
        return {max(d1, d2) + 1, (d1 == d2) ? node : (d1 > d2 ? left.second : right.second)};
    }
};

Github 同步地址:

#865

参考资料:

https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/

https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/discuss/146808/One-pass

https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/discuss/146786/Simple-recursive-Java-Solution

https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/discuss/146842/Short-and-concise-C%2B%2B-solution-using-DFS-3~5-lines

LeetCode All in One 题目讲解汇总(持续更新中...)

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