leetcode98~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.

Example 1:

2

/ \

1 3

Binary tree [2,1,3], return true.

Example 2:

1

/ \

2 3

Binary tree [1,2,3], return false.

方法比较多。

第一种方法比较常规，先中序遍历，然后比较集合中前后值的大小。BST遍历后的序列一定是递增序列。

第二种方法着实比较牛逼了。每次遍历的时候记录遍历时的最大值和最小值。

第三和第四种方法，也是使用中序遍历，不过使用pre指针存储上一次遍历的节点，然后与当前节点进行比较。

public class ValidBST98 {

//先中序遍历，然后比较集合中前后值的大小。BST遍历后的序列一定是递增序列
public boolean isValidBST2(TreeNode root) {
if(root==null)  return true;
TreeNode node = root;
Stack<TreeNode> stack = new Stack<TreeNode>();
List<Integer> list = new ArrayList<Integer>();
while(node!=null || !stack.isEmpty()) {
if(node!=null) {
stack.push(node);
node = node.left;
} else {
TreeNode tmp = stack.pop();
list.add(tmp.val);
node = tmp.right;

}
}

4000
for(int i=0;i<list.size()-1;i++) {
if(list.get(i)>=list.get(i+1)) {
return false;
}
}
return true;
}

//每次遍历的时候记录遍历时的最大值和最小值
public boolean isValidBST3(TreeNode root) {
if(root==null) return true;
if(root.left==null&&root.right==null) return true;
return helper(root,Long.MIN_VALUE,Long.MAX_VALUE);
}

private boolean helper(TreeNode root, long minValue, long maxValue) {
if(root==null) return true;
if(root.val<=minValue || root.val>=maxValue) return false;
return helper(root.left,minValue,root.val) && helper(root.right,root.val,maxValue);
}

//中序遍历 使用pre指针存储上一次遍历的节点，然后与当前节点进行比较
public boolean isValidBST4(TreeNode root) {
if (root == null) return true;
Stack<TreeNode> stack = new Stack<>();
TreeNode pre = null;
while (root != null || !stack.isEmpty()) {
while (root != null) {
stack.push(root);
root = root.left;
}
root = stack.pop();
if(pre != null && root.val <= pre.val) return false;
pre = root;
root = root.right;
}
return true;
}

public boolean isValidBST(TreeNode root){
if(root==null) return true;
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode pre = null;
TreeNode node = root;
while(node!=null || !stack.isEmpty())  {
if(node!=null) {
stack.push(node);
node=node.left;
} else {
TreeNode tmp = stack.pop();
if(pre != null && tmp.val <= pre.val) return false;
pre =tmp;
node = tmp.right;
}
}
return true;
}
}