
Binary Tree Inorder Traversal
The difficult part of inorder traversal is not remembering “left, node, right.” It is preserving the parent node while the left subtree is still…
View solutionDefine what each subtree returns upward or carries downward to solve traversal, validation, aggregation, and structural tree problems.
Problems
Practice problems that share this primary solution pattern and compare the clues that reveal it.

The difficult part of inorder traversal is not remembering “left, node, right.” It is preserving the parent node while the left subtree is still…
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A binary search tree has a structural invariant that is more useful than its parent-child relationships: inorder traversal visits values in sorted order.…
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A tree comparison fails the moment you forget that a missing child is also part of the structure.
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A BST validator must remember the ancestors that still constrain the current node.
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The key to the Symmetric Tree solution is to compare two nodes as a mirror pair, not to traverse the left and right subtrees independently.
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If you visit every node but mix adjacent depths, the traversal is still wrong. The key is to make the queue represent one level at a time.
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The queue does not need to zigzag. It only needs to preserve a predictable order while we place each value into its correct output position.
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The cleanest way to solve this problem is to stop counting depth globally. Ask each subtree for its answer, then let the parent combine those answers.
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The output is bottom-up, but the traversal should still be top-down: collect ordinary BFS levels, then reverse the completed groups.
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A sorted array gives you the middle value immediately. A singly linked list makes you walk for it. The better solution avoids the walk entirely: let…
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A reliable Balanced Binary Tree solution carries two facts upward from every subtree: its height and whether it is balanced.
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The shortest path counts only when it reaches a leaf. That one condition is what breaks the tempting min(left, right) solution.
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A matching sum at an internal node is not enough. The path must start at the root, end at a leaf, and satisfy the target at that boundary.
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That is the central trap in Path Sum II. An internal node may bring the running sum to targetSum, but its path is still incomplete if the node has a child.…
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It places the left subtree in the correct position, but it can disconnect the original right subtree. The real task is to move the left subtree in front of…
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The queue-based solution is easy to see. The constant-space solution is easier to miss: once one level is connected, its next pointers become the queue for…
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The queue solution is easy to derive. The constant-space solution comes from reusing the next pointers you are building as the queue for the next level.
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