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Morris Traversal

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Exuberant Emu answered on April 11, 2023 Popularity 9/10 Helpfulness 3/10

answer Morris Traversal

related morris traversal inorder

Morris Traversal

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inorder morris traversal

1 void inorderMorrisTraversal(TreeNode *root) {
 2     TreeNode *cur = root, *prev = NULL;
 3     while (cur != NULL)
 4     {
 5         if (cur->left == NULL)          // 1.
 6         {
 7             printf("%d ", cur->val);
 8             cur = cur->right;
 9         }
10         else
11         {
12             // find predecessor
13             prev = cur->left;
14             while (prev->right != NULL && prev->right != cur)
15                 prev = prev->right;
16 
17             if (prev->right == NULL)   // 2.a)
18             {
19                 prev->right = cur;
20                 cur = cur->left;
21             }
22             else                       // 2.b)
23             {
24                 prev->right = NULL;
25                 printf("%d ", cur->val);
26                 cur = cur->right;
27             }
28         }
29     }
30 }

Popularity 9/10 Helpfulness 3/10 Language whatever

Source: www.cnblogs.com

Tags: traversal whatever

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Contributed on Apr 11 2023

Exuberant Emu

0 Answers Avg Quality 2/10

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morris traversal inorder

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class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> inorder = new ArrayList<Integer>(); 
        
        TreeNode cur = root; 
        while(cur != null) {
            if(cur.left == null) {
                inorder.add(cur.val); 
                cur = cur.right; 
            }
            else {
                TreeNode prev = cur.left; 
                while(prev.right != null && prev.right != cur) {
                    prev = prev.right; 
                }
                
                if(prev.right == null) {
                    prev.right = cur;
                    cur = cur.left; 
                }
                else {
                    prev.right = null; 
                    inorder.add(cur.val); 
                    cur = cur.right; 
                }
            }
        }
        return inorder; 
    }
}

Popularity 8/10 Helpfulness 1/10 Language java

Source: takeuforward.org

Tags: inorder

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Contributed on Jul 03 2023

012_Aryan Bichpuria

0 Answers Avg Quality 2/10

Morris Traversal

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