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feat(solutions): add recover_binary_search_tree
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‎README.md

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| # | **Problem** & **Solution** | Difficulty | Single Repetition Duration | LeetCode Run Time |
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| ---: | :----- | :----------: | ----------: | ----------: |
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|[99][Solutions-99]|[Recover Binary Search Tree][Solutions-99-Home]|Hard|[94.8 ns/op][Solutions-99-Code] / [3 test cases][Solutions-99-Test]|32 ms|
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|[98][Solutions-98]|[Validate Binary Search Tree][Solutions-98-Home]|Medium|[21.0 ns/op][Solutions-98-Code] / [8 test cases][Solutions-98-Test]|8 ms|
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|[96][Solutions-96]|[Unique Binary Search Trees][Solutions-96-Home]|Medium|[39.1 ns/op][Solutions-96-Code] / [6 test cases][Solutions-96-Test]|0 ms|
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|[95][Solutions-95]|[Unique Binary Search Trees II][Solutions-95-Home]|Medium|[259 ns/op][Solutions-95-Code] / [3 test cases][Solutions-95-Test]|48 ms|
@@ -154,6 +155,10 @@ Welcome to report bugs, suggest ideas and discuss on [issues page](https://githu
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## License
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[![FOSSA Status](https://app.fossa.io/api/projects/git%2Bgithub.com%2FWindomZ%2Fleetcode.go.svg?type=large)](https://app.fossa.io/projects/git%2Bgithub.com%2FWindomZ%2Fleetcode.go?ref=badge_large)
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[Solutions-99]:https://leetcode.com/problems/recover-binary-search-tree/
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[Solutions-99-Home]:solutions/recover_binary_search_tree/
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[Solutions-99-Code]:solutions/recover_binary_search_tree/recoverbinarysearchtree.go
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[Solutions-99-Test]:solutions/recover_binary_search_tree/recoverbinarysearchtree_test.go#L113
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[Solutions-98]:https://leetcode.com/problems/validate-binary-search-tree/
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[Solutions-98-Home]:solutions/validate_binary_search_tree/
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[Solutions-98-Code]:solutions/validate_binary_search_tree/validatebinarysearchtree.go

‎solutions/recover_binary_search_tree/NOTE_Ch-zh.md

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# 解法要点
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1. BST(binary search trees)是左叶值小于节点值，右叶值大于节点值
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1. 运用Morris Traversal方法遍历二叉树，找出两个错误的数值

‎solutions/recover_binary_search_tree/README.md

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# [99. Recover Binary Search Tree](https://leetcode.com/problems/recover-binary-search-tree/description/)
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## Description
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Two elements of a binary search tree (BST) are swapped by mistake.
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Recover the tree without changing its structure.
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**Note:**
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A solution using O(_n_) space is pretty straight forward. Could you devise a constant space solution?
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## Solution
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- [Code](recoverbinarysearchtree.go)
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- [Testing](recoverbinarysearchtree_test.go)
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## Note
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- [中文](NOTE_Ch-zh.md)

‎solutions/recover_binary_search_tree/recoverbinarysearchtree.go

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package recoverbinarysearchtree
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// TreeNode definition for a binary tree node.
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type TreeNode struct {
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Val int
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Left *TreeNode
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Right *TreeNode
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}
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func recoverTree(root *TreeNode) {
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var pre, cur, first, second *TreeNode
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// Morris Traversal
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// https://en.wikipedia.org/wiki/Threaded_binary_tree#The_array_of_Inorder_traversal
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for root != nil {
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if root.Left != nil {
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// connect threading for root
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cur = root.Left
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for cur.Right != nil && cur.Right != root {
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cur = cur.Right
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}
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// the threading already exists
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if cur.Right != nil {
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if pre != nil && pre.Val > root.Val {
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// found the mistaken nodes
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if first == nil {
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first = pre
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}
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second = root
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}
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pre, cur.Right, root = root, nil, root.Right
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} else {
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// construct the threading
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cur.Right, root = root, root.Left
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}
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} else {
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if pre != nil && pre.Val > root.Val {
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// found the mistaken nodes
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if first == nil {
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first = pre
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}
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second = root
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}
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pre, root = root, root.Right
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}
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}
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// swap the two mistaken nodes
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if first != nil && second != nil {
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first.Val, second.Val = second.Val, first.Val
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}
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}

‎solutions/recover_binary_search_tree/recoverbinarysearchtree_test.go

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package recoverbinarysearchtree
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import (
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"testing"
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"github.com/WindomZ/testify/assert"
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)
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func Test_recoverTree(t *testing.T) {
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root := &TreeNode{
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Val: 3,
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Left: &TreeNode{
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Val: 1,
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Left: &TreeNode{Val: 2},
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},
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}
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recoverTree(root)
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assert.Equal(t, &TreeNode{
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Val: 3,
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Left: &TreeNode{
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Val: 2,
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Left: &TreeNode{Val: 1},
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},
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}, root)
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root = &TreeNode{
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Val: 2,
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Left: &TreeNode{
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Val: 3,
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},
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Right: &TreeNode{
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Val: 1,
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},
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}
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recoverTree(root)
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assert.Equal(t, &TreeNode{
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Val: 2,
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Left: &TreeNode{
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Val: 1,
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},
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Right: &TreeNode{
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Val: 3,
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},
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}, root)
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root = &TreeNode{
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Val: 3,
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Right: &TreeNode{
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Val: 2,
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Right: &TreeNode{
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Val: 1,
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},
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},
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}
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recoverTree(root)
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assert.Equal(t, &TreeNode{
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Val: 1,
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Right: &TreeNode{
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Val: 2,
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Right: &TreeNode{
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Val: 3,
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},
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},
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}, root)
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root = &TreeNode{
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Val: 4,
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Left: &TreeNode{
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Val: 2,
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Left: &TreeNode{Val: 3},
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},
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Right: &TreeNode{Val: 5},
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}
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recoverTree(root)
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assert.Equal(t, &TreeNode{
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Val: 4,
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Left: &TreeNode{
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Val: 3,
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Left: &TreeNode{Val: 2},
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},
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Right: &TreeNode{Val: 5},
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}, root)
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root = &TreeNode{
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Val: 4,
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Left: &TreeNode{
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Val: 2,
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Right: &TreeNode{
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Val: 3,
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}},
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Right: &TreeNode{
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Val: 6,
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Left: &TreeNode{
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Val: 7,
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}},
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}
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recoverTree(root)
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assert.Equal(t, &TreeNode{
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Val: 4,
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Left: &TreeNode{
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Val: 2,
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Right: &TreeNode{
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Val: 3,
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}},
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Right: &TreeNode{
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Val: 7,
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Left: &TreeNode{
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Val: 6,
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}},
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}, root)
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}
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func Benchmark_recoverTree(b *testing.B) {
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b.StopTimer()
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b.ReportAllocs()
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b.StartTimer()
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b.RunParallel(func(pb *testing.PB) {
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for pb.Next() {
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recoverTree(&TreeNode{
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Val: 3,
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Left: &TreeNode{
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Val: 1,
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Left: &TreeNode{Val: 2},
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},
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})
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recoverTree(&TreeNode{
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Val: 1,
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Left: &TreeNode{Val: 2},
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Right: &TreeNode{Val: 3},
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})
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recoverTree(&TreeNode{
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Val: 3,
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Left: &TreeNode{
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Val: 2,
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Right: &TreeNode{
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Val: 4,
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}},
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Right: &TreeNode{
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Val: 3,
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Left: &TreeNode{
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Val: 4,
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}},
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})
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}
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})
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}

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