Same Tree

Same Tree - Problem

Given the roots of two binary trees p and q, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.

Input & Output

Example 1 — Identical Trees

$ Input: p = [1,2,3], q = [1,2,3]

› Output: true

💡 Note: Both trees have the same structure: root 1 with left child 2 and right child 3. All corresponding nodes have identical values.

Example 2 — Different Structure

$ Input: p = [1,2], q = [1,null,2]

› Output: false

💡 Note: Tree p has node 2 as left child of root 1, while tree q has node 2 as right child of root 1. Different structure makes them not identical.

Example 3 — Different Values

$ Input: p = [1,2,1], q = [1,1,2]

› Output: false

💡 Note: Same structure but different values: p has left=2, right=1 while q has left=1, right=2. Values don't match at corresponding positions.

Constraints

The number of nodes in both trees is in the range [0, 100]
-10⁴ ≤ Node.val ≤ 10⁴

Visualization

Tap to expand

Asked in

G Google 35 a Amazon 28 M Microsoft 22 🍎 Apple 15

The key insight is that two trees are identical if their roots match AND their left/right subtrees are also identical. Best approach is recursive DFS which elegantly handles this by checking current nodes then recursing on children. Time: O(min(m,n)), Space: O(min(m,n))

Common Approaches

✓ Breadth-First Search (Level Order)

⏱️ Time: O(min(m,n)) Space: O(min(m,n))

Use two queues to traverse both trees simultaneously in level order. Compare nodes at each level to ensure structure and values match.

Recursive DFS Comparison

⏱️ Time: O(min(m,n)) Space: O(min(m,n))

Check if current nodes have same value, then recursively check left and right subtrees. If any mismatch found, trees are different.

Breadth-First Search (Level Order) — Algorithm Steps

Initialize queues with root nodes
Process nodes level by level
Compare values and add children to queues

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize Queues

Add root nodes to both queues

Process Level

Compare nodes at current level

Add Children

Add children to queues for next level

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
};

bool solution(struct TreeNode* p, struct TreeNode* q) {
    if (!p && !q) return true;
    if (!p || !q) return false;
    
    struct TreeNode* queueP[10000];
    struct TreeNode* queueQ[10000];
    int frontP = 0, rearP = 0, frontQ = 0, rearQ = 0;
    
    queueP[rearP++] = p;
    queueQ[rearQ++] = q;
    
    while (frontP < rearP && frontQ < rearQ) {
        struct TreeNode* nodeP = queueP[frontP++];
        struct TreeNode* nodeQ = queueQ[frontQ++];
        
        if (!nodeP && !nodeQ) continue;
        if (!nodeP || !nodeQ) return false;
        if (nodeP->val != nodeQ->val) return false;
        
        queueP[rearP++] = nodeP->left;
        queueP[rearP++] = nodeP->right;
        queueQ[rearQ++] = nodeQ->left;
        queueQ[rearQ++] = nodeQ->right;
    }
    
    return (rearP - frontP) == (rearQ - frontQ);
}

struct TreeNode* createNode(int val) {
    struct TreeNode* node = malloc(sizeof(struct TreeNode));
    node->val = val;
    node->left = NULL;
    node->right = NULL;
    return node;
}

struct TreeNode* buildTree(int* arr, int size, int index) {
    if (index >= size || arr[index] == -1001) return NULL;
    struct TreeNode* root = createNode(arr[index]);
    root->left = buildTree(arr, size, 2 * index + 1);
    root->right = buildTree(arr, size, 2 * index + 2);
    return root;
}

int parseArray(const char* str, int* arr) {
    int size = 0;
    const char* p = str;
    
    // Find opening bracket
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    // Parse elements
    while (*p && *p != ']') {
        // Skip whitespace and commas
        while (*p == ' ' || *p == ',') p++;
        
        if (*p == ']' || *p == '\0') break;
        
        // Check for null
        if (strncmp(p, "null", 4) == 0) {
            arr[size++] = -1001;  // Sentinel for null
            p += 4;
        } else {
            // Parse number
            char* endptr;
            arr[size++] = (int)strtol(p, &endptr, 10);
            p = endptr;
        }
    }
    
    return size;
}

int main() {
    char line1[1000], line2[1000];
    fgets(line1, sizeof(line1), stdin);
    fgets(line2, sizeof(line2), stdin);
    
    int pArr[1000], qArr[1000];
    int pSize = parseArray(line1, pArr);
    int qSize = parseArray(line2, qArr);
    
    struct TreeNode* p = buildTree(pArr, pSize, 0);
    struct TreeNode* q = buildTree(qArr, qSize, 0);
    
    printf(solution(p, q) ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(min(m,n))

Visit each node once until mismatch found, where m and n are tree sizes

✓ Linear Growth

Space Complexity

O(min(m,n))

Queues store nodes at current level, worst case is full bottom level

⚡ Linearithmic Space

312.0K Views

High Frequency

~15 min Avg. Time

2.8K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Breadth-First Search (Level Order) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler