Second Place - Amazon Top Interview Questions

Problem Statement :

Given a binary tree root, return the depth of the second deepest leaf. Note that if there are multiple deepest leaves, the second deepest leaf is the next highest one.

The root has a depth of 0 and you can assume the answer is guaranteed to exist for the trees given.


n ≤ 100,000 where n is the number of nodes in root.

Example 1


root = [1, [2, null, null], [3, [4, [6, null, null], null], [5, null, [7, null, null]]]]




The the second deepest leaf is 2 which exists at depth 1

Solution :


                        Solution in C++ :

int solve(Tree* root) {
    queue<Tree*> q;

    int depth = 0;
    int deepest = 0;
    int res = 0;
    while (!q.empty()) {
        int n = q.size();
        while (n--) {
            Tree* t = q.front();

            if (t->left == t->right) {
                if (depth > deepest) {
                    res = deepest;
                    deepest = depth;
                } else if (depth > res and depth < deepest)
                    res = depth;

            if (t->left) q.push(t->left);
            if (t->right) q.push(t->right);

    return res;

                        Solution in Java :

import java.util.*;

 * public class Tree {
 *   int val;
 *   Tree left;
 *   Tree right;
 * }
class Solution {
    int max1 = -1, max2 = -1;

    public void dfs(Tree root, int depth) {
        if (root == null)
        this.dfs(root.left, depth + 1);
        this.dfs(root.right, depth + 1);
        if (root.left == null && root.right == null)
            if (depth == max1)
            else if (depth > max1) {
                max2 = max1;
                max1 = depth;
            } else if (depth > max2)
                max2 = depth;

    public int solve(Tree root) {
        dfs(root, 0);
        return max2;

                        Solution in Python : 
class Solution:
    def solve(self, root):
        depths = set()

        def dfs(node, d=0):
            if node:
                if not node.left and not node.right:
                dfs(node.left, d + 1)
                dfs(node.right, d + 1)

        return max(depths)

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