Recover Binary Search Tree(恢復二叉搜尋樹)
阿新 • • 發佈:2019-01-03
題目原型:
Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Note:A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?
基本思路:
這個題的意思就是一個二叉查詢樹有兩個元素被調換了順序,此時需要修復成二叉查詢樹。基於此,由於構成二叉查詢樹的序列是二叉樹的中序遍歷序列,所以一旦我們根據中序遍歷找到兩個“變動”的點然後交換他們的值即可。
TreeNode pre = null;//始終指向中序遍歷的上一個節點 TreeNode n1 = null;//記錄待交換節點 TreeNode n2 = null;//記錄帶交換節點 public TreeNode recoverTree(TreeNode root) { findTwoNode(root); if(n1!=null&&n2!=null) { int tmp = n1.val; n1.val = n2.val; n2.val = tmp; } return root; } public void findTwoNode(TreeNode root) { if(root==null) return; findTwoNode(root.left); if(pre!=null&&pre.val>root.val) { if(n1==null) n1 = pre; n2 = root; } pre = root; findTwoNode(root.right); }