1 """ 2 Two elements of a binary search tree (BST) are swapped by mistake. 3 Recover the tree without changing its structure. 4 Example 1: 5 Input: [1,3,null,null,2] 6 7 1 8 / 9 3 10 11 2 12 Output: [3,1,null,null,2] 13 3 14 / 15 1 16 17 2 18 Example 2: 19 Input: [3,1,4,null,null,2] 20 3 21 / 22 1 4 23 / 24 2 25 Output: [2,1,4,null,null,3] 26 2 27 / 28 1 4 29 / 30 3 31 """ 32 """ 33 自己AC了一种笨方法： 34 首先对二叉树中序遍历得到inorder 35 再对inorder排序得到s_inorder 36 比较两个列表的差异，如果不相等，将两个不相等的值保存到x,y 37 第二次遍历二叉树，将值为x的设为'a'，值为y的设为x。(这里不知道怎么交换值) 38 第三次遍历二叉树，将值为'a'的设为y 39 """ 40 class TreeNode: 41 def __init__(self, x): 42 self.val = x 43 self.left = None 44 self.right = None 45 46 class Solution1: 47 def recoverTree(self, root): 48 """ 49 Do not return anything, modify root in-place instead. 50 """ 51 stack = [] 52 cur = root 53 inorder = [] 54 while cur or stack: 55 if cur: 56 stack.append(cur) 57 cur = cur.left 58 else: 59 cur = stack.pop() 60 inorder.append(cur.val) 61 cur = cur.right 62 s_inorder = sorted(inorder) 63 for i in range(len(s_inorder)): 64 if s_inorder[i] != inorder[i]: 65 x = s_inorder[i] 66 y = inorder[i] 67 break 68 queue = [] 69 queue.append(root) 70 while queue: 71 node = queue.pop(0) 72 if node.val == x: 73 node.val = 'a' 74 if node.val == y: 75 node.val = x 76 if node.left: 77 queue.append(node.left) 78 if node.right: 79 queue.append(node.right) 80 newqueue = [] 81 newqueue.append(root) 82 while newqueue: 83 _node = newqueue.pop(0) 84 if _node.val == 'a': 85 _node.val = y 86 if _node.left: 87 newqueue.append(_node.left) 88 if _node.right: 89 newqueue.append(_node.right) 90 """ 91 解法二：可以再解法一得到有序的s_inorder后 92 将s_inorder覆盖到整个二叉树 93 """ 94 class Solution2: 95 def recoverTree(self, root): 96 """ 97 Do not return anything, modify root in-place instead. 98 """ 99 stack = [] 100 cur = root 101 inorder = [] 102 while cur or stack: 103 if cur: 104 stack.append(cur) 105 cur = cur.left 106 else: 107 cur = stack.pop() 108 inorder.append(cur.val) 109 cur = cur.right 110 s_inorder = sorted(inorder) 111 i = 0 112 _cur = root 113 _stack = [] 114 while _cur or _stack: 115 if _cur: 116 _stack.append(_cur) 117 _cur = _cur.left 118 else: 119 _cur = _stack.pop() 120 _cur.val = s_inorder[i] 121 i += 1 122 _cur = _cur.right 123 124 if __name__ == '__main__': 125 root = TreeNode(1) 126 node1 = TreeNode(3) 127 node2 = TreeNode(2) 128 root.left = node1 129 node1.right = node2 130 ans = Solution1() 131 print(ans.recoverTree(root))