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Leetcode Binary Tree Zigzag Level Order Traversal problem solution

YASH PAL, 31 July 2024

In this Leetcode Binary Tree Zigzag Level Order Traversal problem solution we have Given the root of a binary tree, return the zigzag level order traversal of its nodes’ values.

Leetcode Binary Tree Zigzag Level Order Traversal problem solution

Problem solution in Python.

class Solution:
    def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]:
        
        to_right = 1
        
        if root == None:
            return []
        
        cur_queue = [root]
        next_queue = []
        result = []
        tmp = []
        while(cur_queue):
            
            node = cur_queue.pop()
            tmp.append(node.val)
            if(to_right):
                if(node.left):
                    next_queue.append(node.left)
                if(node.right):
                    next_queue.append(node.right)
            else:
                if(node.right):
                    next_queue.append(node.right)
                if(node.left):
                    next_queue.append(node.left)
            
            if(not cur_queue):
                cur_queue = next_queue
                next_queue = []
                to_right = not to_right
                result.append(tmp)
                tmp = []
        
        return result

Problem solution in Java.

public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root==null) return res;
        Queue<TreeNode> q = new LinkedList<>();
        boolean reverse=false;
        q.offer(root);
        while(!q.isEmpty()){
            int l = q.size(); 
            List<Integer> ll = new LinkedList<>();

            for(int i=0; i < l ; i++){
                
                TreeNode ln = q.poll();
                if(reverse) ll.add(0,ln.val);
                else ll.add(ln.val);
                if(ln.left!=null) q.offer(ln.left);
                if(ln.right!=null) q.offer(ln.right);
            
            }
            res.add(ll);
            reverse = reverse?false:true;
        }
        return res;
        
    }

Problem solution in C++.

vector<vector<int>> ans;
class Solution {
public:
    void helper(TreeNode* root,int level=0)
    {
        if(root==NULL)
            return;
        if(level>=ans.size())
        {
            ans.resize(level+1);
        }
        if(level&1)
        {
            ans[level].insert(ans[level].begin()+0,root->val);
        }
        else{
            ans[level].push_back(root->val);
        }
        helper(root->left,level+1);
        helper(root->right,level+1);
    }
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        ans.clear();
        ans.resize(0);
        helper(root);
        return ans;
    }
};

Problem solution in C.

#include<math.h>
int max (int a, int b)
{
    return (a>b?a:b);
}

int findHieght(struct TreeNode *root) 
{
    if(root == NULL) return 0;
    int lt = findHieght(root->left);
    int rt = findHieght(root->right);
    return 1 + max(lt,rt);
}

int** zigzagLevelOrder(struct TreeNode* root, int* returnSize, int** returnColumnSizes){

    int **sol;
    int ht = findHieght(root), currLevel;
    struct TreeNode* s1[1024];
    struct TreeNode* s2[1024];
    struct TreeNode* tmp;
    int top1, top2, col;
    int i, j;
    
    top1 = top2 = -1;
    *returnSize = ht;
    if (root == NULL) return NULL;
    
    sol = (int **)malloc(sizeof(int *)*ht);
    *returnColumnSizes = (int *)malloc (sizeof(int) * ht);
    s1[++top1] = root;
    currLevel = -1;
    col = 0;
    while (top1 != -1 || top2 != -1) {
        if (top1 != -1) {
            sol[++currLevel] = (int *)malloc(sizeof(int)*(top1+1));
            col = 0;
        }
        while(top1 != -1) {
            tmp = s1[top1--];
            if (tmp->left) s2[++top2] = tmp->left;
            if (tmp->right) s2[++top2] = tmp->right;
            sol[currLevel][col++] = tmp->val;
            (*returnColumnSizes)[currLevel] = col; /* 1 based indexing */
            
        }
        
        if (top2 != -1) {
            sol[++currLevel] = (int *)malloc(sizeof(int)*(top2+1));
            col = 0;
        }
        while(top2 != -1) {
            tmp = s2[top2--];
            if (tmp->right) s1[++top1] = tmp->right;
            if (tmp->left) s1[++top1] = tmp->left;
            sol[currLevel][col++] = tmp->val;
            (*returnColumnSizes)[currLevel] = col;
        }
    }
    *returnSize = ht;
    #if 0
    for (i=0;i<ht;i++) {
        for (j=0;j<(*returnColumnSizes)[i]; j++) {
            printf("%d ", sol[i][j]);
        }
    }
    #endif
    return sol;
}

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