Question

1123. Lowest Common Ancestor of Deepest Leaves

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Description

Given the root of a binary tree, return _the lowest common ancestor of its deepest leaves_.

Recall that:

• The node of a binary tree is a leaf if and only if it has no children
• The depth of the root of the tree is 0. if the depth of a node is d, the depth of each of its children is d + 1.
• The lowest common ancestor of a set S of nodes, is the node A with the largest depth such that every node in S is in the subtree with root A.

 

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 leaf-nodes of the tree.
Note that nodes 6, 0, and 8 are also leaf nodes, but the depth of them is 2, but the depth of nodes 7 and 4 is 3.

Example 2:

Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree, and it's the lca of itself.

Example 3:

Input: root = [0,1,3,null,2]
Output: [2]
Explanation: The deepest leaf node in the tree is 2, the lca of one node is itself.

 

Constraints:

• The number of nodes in the tree will be in the range [1, 1000].
• 0 <= Node.val <= 1000
• The values of the nodes in the tree are unique.

 

Note: This question is the same as 865: https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/

Solutions

Solution 1: DFS

We design a function dfs(root) that returns a tuple (l, d), where l is the deepest common ancestor of node root, and d is the depth of node root. The execution logic of the function dfs(root) is as follows:

• If root is null, return the tuple (None, 0);
• Otherwise, we recursively call dfs(root.left) and dfs(root.right), obtaining tuples (l, d1) and (r, d2). If d1 > d2, the deepest common ancestor of root is l, and the depth is d1 + 1; if d1 < d2, the deepest common ancestor of root is r, and the depth is d2 + 1; if d1 = d2, the deepest common ancestor of root is root, and the depth is d1 + 1.

In the main function, we call dfs(root) and return the first element of its return value to get the deepest common ancestor node.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def lcaDeepestLeaves(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
    def dfs(root):
      if root is None:
        return None, 0
      l, d1 = dfs(root.left)
      r, d2 = dfs(root.right)
      if d1 > d2:
        return l, d1 + 1
      if d1 < d2:
        return r, d2 + 1
      return root, d1 + 1

    return dfs(root)[0]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  public TreeNode lcaDeepestLeaves(TreeNode root) {
    return dfs(root).getKey();
  }

  private Pair<TreeNode, Integer> dfs(TreeNode root) {
    if (root == null) {
      return new Pair<>(null, 0);
    }
    Pair<TreeNode, Integer> l = dfs(root.left);
    Pair<TreeNode, Integer> r = dfs(root.right);
    int d1 = l.getValue(), d2 = r.getValue();
    if (d1 > d2) {
      return new Pair<>(l.getKey(), d1 + 1);
    }
    if (d1 < d2) {
      return new Pair<>(r.getKey(), d2 + 1);
    }
    return new Pair<>(root, d1 + 1);
  }
}

/**
 * Definition for a binary tree node.
 * 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) {}
 * };
 */
class Solution {
public:
  TreeNode* lcaDeepestLeaves(TreeNode* root) {
    return dfs(root).first;
  }

  pair<TreeNode*, int> dfs(TreeNode* root) {
    if (!root) {
      return {nullptr, 0};
    }
    auto [l, d1] = dfs(root->left);
    auto [r, d2] = dfs(root->right);
    if (d1 > d2) {
      return {l, d1 + 1};
    }
    if (d1 < d2) {
      return {r, d2 + 1};
    }
    return {root, d1 + 1};
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
type pair struct {
  first  *TreeNode
  second int
}

func lcaDeepestLeaves(root *TreeNode) *TreeNode {
  var dfs func(root *TreeNode) pair
  dfs = func(root *TreeNode) pair {
    if root == nil {
      return pair{nil, 0}
    }
    l, r := dfs(root.Left), dfs(root.Right)
    d1, d2 := l.second, r.second
    if d1 > d2 {
      return pair{l.first, d1 + 1}
    }
    if d1 < d2 {
      return pair{r.first, d2 + 1}
    }
    return pair{root, d1 + 1}
  }
  return dfs(root).first
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function lcaDeepestLeaves(root: TreeNode | null): TreeNode | null {
  const dfs = (root: TreeNode | null): [TreeNode | null, number] => {
    if (root === null) {
      return [null, 0];
    }
    const [l, d1] = dfs(root.left);
    const [r, d2] = dfs(root.right);
    if (d1 > d2) {
      return [l, d1 + 1];
    }
    if (d1 < d2) {
      return [r, d2 + 1];
    }
    return [root, d1 + 1];
  };
  return dfs(root)[0];
}