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865. Smallest Subtree with all the Deepest Nodes (Medium)

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

Return the smallest subtree such that it contains all the deepest nodes in the original tree.

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

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

Note: This question is the same as 1123: https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/

 

Example 1:

Input: root = [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 coloured in blue are the deepest nodes of the tree.
Notice that nodes 5, 3 and 2 contain the deepest nodes in the tree but node 2 is the smallest subtree among them, so we return it.

Example 2:

Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree.

Example 3:

Input: root = [0,1,3,null,2]
Output: [2]
Explanation: The deepest node in the tree is 2, the valid subtrees are the subtrees of nodes 2, 1 and 0 but the subtree of node 2 is the smallest.

 

Constraints:

  • The number of nodes in the tree will be in the range [1, 500].
  • 0 <= Node.val <= 500
  • The values of the nodes in the tree are unique.

Related Topics:
Tree, Depth-first Search, Breadth-first Search, Recursion

Solution 1.

## Solution 1.

```cpp
// OJ: https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
    int maxDepth = -1, target = 0, cnt = 0;
    void count(TreeNode *root, int d) {
        if (!root) return;
        if (d > maxDepth) {
            target = 1;
            maxDepth = d;
        } else if (d == maxDepth) ++target;
        count(root->left, d + 1);
        count(root->right, d + 1);
    }
    TreeNode *find(TreeNode *root, int d) {
        if (!root) return NULL;
        int before = cnt;
        auto left = find(root->left, d + 1);
        if (left) return left;
        auto right = find(root->right, d + 1);
        if (right) return right;
        if (d == maxDepth) ++cnt;
        return before == 0 && cnt == target ? root : NULL;
    }
public:
    TreeNode* lcaDeepestLeaves(TreeNode* root) {
        count(root, 0);
        return find(root, 0);
    }
};

Solution 2.

The lowest ancester is the highest node whose left and right subtrees have the same height.

// OJ: https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
    pair<TreeNode*, int> dfs(TreeNode *root, int d = 0) { // latest node which has equal depth in left and right sub-trees; the corresponding height
        if (!root) return {NULL, 0};
        const auto &[left, ld] = dfs(root->left, d + 1);
        const auto &[right, rd] = dfs(root->right, d + 1);
        if (ld > rd) return {left, ld + 1};
        else if (ld < rd) return{right, rd + 1};
        return {root, ld + 1};
    }
public:
    TreeNode* lcaDeepestLeaves(TreeNode* root) {
        return dfs(root).first;
    }
};