LeetCode: Validate Binary Search Tree 解题报告
Validate Binary Search Tree
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
SOLUTION 1:
使用Iterator 中序遍历的方法,判断整个数列是否保持增序即可。
算法思想:
http://www.cnblogs.com/shuaiwhu/archive/2011/04/20/2065055.html
1.采用栈的话,先寻找最左边的节点,把经过的节点都存入栈中,第一个被弹出来的为最左节点,那么访问其右子树,对右子树也像前面一样遍历,整个流程跟递归一样。
1 public boolean isValidBST1(TreeNode root) { 2 // Just use the inOrder traversal to solve the problem. 3 if (root == null) { 4 return true; 5 } 6 7 Stack<TreeNode> s = new Stack<TreeNode>(); 8 TreeNode cur = root; 9 10 TreeNode pre = null; 11 12 while(true) { 13 // Push all the left node into the stack. 14 while (cur != null) { 15 s.push(cur); 16 cur = cur.left; 17 } 18 19 if (s.isEmpty()) { 20 break; 21 } 22 23 // No left node, just deal with the current node. 24 cur = s.pop(); 25 26 if (pre != null && pre.val >= cur.val) { 27 return false; 28 } 29 30 pre = cur; 31 32 // Go to the right node. 33 cur = cur.right; 34 } 35 36 return true; 37 }
SOLUTION 2:
引自大神的思想:http://blog.csdn.net/fightforyourdream/article/details/14444883
我们可以设置上下bound,递归左右子树时,为它们设置最大值,最小值,并且不可以超过。
注意:下一层递归时,需要把本层的up 或是down继续传递下去。相当巧妙的算法。
1 /* 2 SOLUTION 2: Use the recursive version. 3 REF: http://blog.csdn.net/fightforyourdream/article/details/14444883 4 */ 5 public boolean isValidBST2(TreeNode root) { 6 // Just use the inOrder traversal to solve the problem. 7 if (root == null) { 8 return true; 9 } 10 11 return dfs(root, Long.MIN_VALUE, Long.MAX_VALUE); 12 } 13 14 public boolean dfs(TreeNode root, long low, long up) { 15 if (root == null) { 16 return true; 17 } 18 19 if (root.val >= up || root.val <= low) { 20 return false; 21 } 22 23 return dfs(root.left, low, root.val) 24 && dfs(root.right, root.val, up); 25 }
SOLUTION 3:
同样是递归,但是把左右子树的min, max值返回,与当前的root值相比较。比较直观。
1 /* 2 SOLUTION 3: Use the recursive version2. 3 */ 4 public boolean isValidBST3(TreeNode root) { 5 // Just use the inOrder traversal to solve the problem. 6 if (root == null) { 7 return true; 8 } 9 10 return dfs(root, Long.MIN_VALUE, Long.MAX_VALUE); 11 } 12 13 public class ReturnType { 14 int min; 15 int max; 16 boolean isBST; 17 public ReturnType (int min, int max, boolean isBST) { 18 this.min = min; 19 this.max = max; 20 this.isBST = isBST; 21 } 22 } 23 24 // BST: 25 // 1. Left tree is BST; 26 // 2. Right tree is BST; 27 // 3. root value is bigger than the max value of left tree and 28 // smaller than the min value of the right tree. 29 public ReturnType dfs(TreeNode root) { 30 ReturnType ret = new ReturnType(Integer.MAX_VALUE, Integer.MIN_VALUE, true); 31 if (root == null) { 32 return ret; 33 } 34 35 ReturnType left = dfs(root.left); 36 ReturnType right = dfs(root.right); 37 38 // determine the left tree and the right tree; 39 if (!left.isBST || !right.isBST) { 40 ret.isBST = false; 41 return ret; 42 } 43 44 // 判断Root.left != null是有必要的,如果root.val是MAX 或是MIN value,判断会失误 45 if (root.left != null && root.val <= left.max) { 46 ret.isBST = false; 47 return ret; 48 } 49 50 if (root.right != null && root.val >= right.min) { 51 ret.isBST = false; 52 return ret; 53 } 54 55 return new ReturnType(Math.min(root.val, left.min), Math.max(root.val, right.max), true); 56 }
SOLUTION 4:
使用一个全局变量,用递归的中序遍历来做,也很简单(但全局变量主页君不推荐!)
1 /* 2 SOLUTION 4: Use the recursive version3. 3 */ 4 TreeNode pre = null; 5 6 public boolean isValidBST(TreeNode root) { 7 // Just use the inOrder traversal to solve the problem. 8 return dfs4(root); 9 } 10 11 public boolean dfs4(TreeNode root) { 12 if (root == null) { 13 return true; 14 } 15 16 // Judge the left tree. 17 if (!dfs4(root.left)) { 18 return false; 19 } 20 21 // judge the sequence. 22 if (pre != null && root.val <= pre.val) { 23 return false; 24 } 25 pre = root; 26 27 // Judge the right tree. 28 if (!dfs4(root.right)) { 29 return false; 30 } 31 32 return true; 33 }
GITHUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/IsValidBST_1221_2014.java
posted on 2014-12-21 21:08 Yu's Garden 阅读(7808) 评论(0) 编辑 收藏 举报