Delete Leaves With a Given Value

Delete Leaves With a Given Value - Problem

Given a binary tree root and an integer target, delete all the leaf nodes with value target.

Note: Once you delete a leaf node with value target, if its parent node becomes a leaf node and has the value target, it should also be deleted. You need to continue doing this until you cannot delete any more nodes.

Input & Output

Example 1 — Basic Leaf Deletion

$ Input: root = [1,2,3,2,null,2,4], target = 2

› Output: [1,null,3,null,4]

💡 Note: First remove leaf nodes with value 2, which are at positions. After removal, some nodes become new leaves with value 2, so continue deleting until no target leaves remain.

Example 2 — Cascade Deletion

$ Input: root = [1,3,3,3,2], target = 3

› Output: [1,3,null,null,2]

💡 Note: Remove leaf node 3, then the parent 3 becomes a leaf and gets removed too. Continue until no more target leaves exist.

Example 3 — Complete Deletion

$ Input: root = [1,1,1], target = 1

› Output: []

💡 Note: All leaf nodes are 1, after removing them, root becomes leaf with value 1, so it gets deleted too. Final tree is empty.

Constraints

1 ≤ Number of nodes ≤ 3000
1 ≤ Node.val ≤ 1000
1 ≤ target ≤ 1000

Visualization

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Asked in

a Amazon 15 f Facebook 12 M Microsoft 8

The key insight is using post-order DFS traversal to process children before parents, allowing single-pass deletion of target leaf nodes. Best approach is Post-order DFS with Time: O(n), Space: O(h).

Common Approaches

✓ Multiple Pass Traversal

⏱️ Time: O(n²) Space: O(h)

Keep scanning the entire tree multiple times, removing target leaf nodes in each pass. Continue until a complete pass finds no nodes to delete.

Post-order DFS (Optimal)

⏱️ Time: O(n) Space: O(h)

Traverse the tree in post-order (left, right, root), deleting target leaf nodes. Since we process children first, when we reach a parent, we already know if its children were deleted.

Multiple Pass Traversal — Algorithm Steps

Step 1: Traverse tree and identify all leaf nodes with target value
Step 2: Delete found target leaves
Step 3: Repeat until no target leaves remain

Visualization

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Step-by-Step Walkthrough

First Pass

Remove all target leaves

Second Pass

New leaves exposed, remove target ones

Final Tree

No more target leaves remain

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;
};

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

struct TreeNode* arrayToTree(char tokens[][20], int size) {
    if (size == 0 || strcmp(tokens[0], "null") == 0) return NULL;
    
    struct TreeNode* root = createNode(atoi(tokens[0]));
    struct TreeNode* queue[1000];
    int front = 0, rear = 0;
    queue[rear++] = root;
    int i = 1;
    
    while (front < rear && i < size) {
        struct TreeNode* node = queue[front++];
        
        if (i < size && strcmp(tokens[i], "null") != 0) {
            node->left = createNode(atoi(tokens[i]));
            queue[rear++] = node->left;
        }
        i++;
        
        if (i < size && strcmp(tokens[i], "null") != 0) {
            node->right = createNode(atoi(tokens[i]));
            queue[rear++] = node->right;
        }
        i++;
    }
    return root;
}

int treeToArray(struct TreeNode* node, char result[][20]) {
    if (!node) return 0;
    
    struct TreeNode* queue[1000];
    int front = 0, rear = 0;
    queue[rear++] = node;
    int count = 0;
    
    while (front < rear) {
        struct TreeNode* current = queue[front++];
        
        if (current) {
            sprintf(result[count++], "%d", current->val);
            queue[rear++] = current->left;
            queue[rear++] = current->right;
        } else {
            strcpy(result[count++], "null");
        }
    }
    
    while (count > 0 && strcmp(result[count-1], "null") == 0) {
        count--;
    }
    
    return count;
}

struct TreeNode* dfs(struct TreeNode* node, int target) {
    if (!node) return NULL;
    
    // Post-order: process children first
    node->left = dfs(node->left, target);
    node->right = dfs(node->right, target);
    
    // If current node is now a leaf with target value, delete it
    if (!node->left && !node->right && node->val == target) {
        free(node);
        return NULL;
    }
    
    return node;
}

int solution(struct TreeNode* root, int target, char result[][20]) {
    root = dfs(root, target);
    return treeToArray(root, result);
}

int parseArray(const char* line, char tokens[][20]) {
    int count = 0;
    int len = strlen(line);
    if (len < 3) return 0; // "[]"
    
    char* content = (char*)malloc(len - 1);
    strncpy(content, line + 1, len - 2); // Remove [ ]
    content[len - 2] = '\0';
    
    if (strlen(content) == 0) {
        free(content);
        return 0;
    }
    
    char* token = strtok(content, ",");
    while (token != NULL && count < 1000) {
        // Trim whitespace
        while (*token == ' ' || *token == '\t') token++;
        char* end = token + strlen(token) - 1;
        while (end > token && (*end == ' ' || *end == '\t')) end--;
        *(end + 1) = '\0';
        
        strcpy(tokens[count++], token);
        token = strtok(NULL, ",");
    }
    
    free(content);
    return count;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = '\0'; // Remove newline
    
    int target;
    scanf("%d", &target);
    
    char tokens[1000][20];
    int tokenCount = parseArray(line, tokens);
    
    struct TreeNode* root = arrayToTree(tokens, tokenCount);
    
    char result[1000][20];
    int resultSize = solution(root, target, result);
    
    printf("[");
    for (int i = 0; i < resultSize; i++) {
        if (i > 0) printf(",");
        if (strcmp(result[i], "null") == 0) {
            printf("null");
        } else {
            printf("%s", result[i]);
        }
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

In worst case, need O(n) passes, each taking O(n) time

⚠ Quadratic Growth

Space Complexity

O(h)

Recursion stack depth equals tree height h

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Multiple Pass Traversal — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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