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We run a preorder depth-first search (DFS) on the root of a binary tree. At each node in this traversal, we output D dashes (where D is the depth of this node), then we output the value of this node. If the depth of a node is D, the depth of its immediate child is D + 1. The depth of the root node is 0. If a node has only one child, that child is guaranteed to be the left child. Given the output traversal of this traversal, recover the tree and return its root.
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We run a preorder depth-first search (DFS) on the root of a binary tree. At each node in this traversal, we output D dashes (where D is the depth of this node), then we output the value of this node. If the depth of a node is D, the depth of its immediate child is D + 1. The depth of the root node is 0. If a node has only one child, that child is guaranteed to be the left child. Given the output traversal of this traversal, recover the tree and return its root.
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traversal = "1-2--3--4-5--6--7"
[1,2,5,3,4,6,7]
traversal = "1-2--3---4-5--6---7"
[1,2,5,3,null,6,null,4,null,7]
traversal = "1-401--349---90--88"
[1,401,null,349,88,90]
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Solve Recover a Tree From Preorder Traversal — We run a preorder depth-first search (DFS) on the root of a binary tree. At each...
Here's the optimal approach using String:
Time: O(n) | Space: O(n)
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