题目:

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.

给出一棵二叉树,判断是否符合给定的规则

思路:

考虑二叉树的中序遍历。

二叉树的中序遍历是升序的,因此中序遍历后判断序列是否是升序即可。

代码:

    public static boolean isValidBST(TreeNode root){
if(root == null){
return true;
}
Stack<TreeNode> ts = new Stack<TreeNode>();
ArrayList<Integer> arr = new ArrayList<Integer>();
TreeNode ptr = root;
//中序遍历
while(ptr != null || !ts.isEmpty()){
while(ptr != null){
ts.push(ptr);
ptr = ptr.left;
} if( !ts.isEmpty() ){
ptr = ts.pop();
arr.add(ptr.val);
ptr = ptr.right;
}
}
int index = 0;
int[] arrs = new int[arr.size()];
for(int i: arr){
arrs[index++] = i;
}
//判断中序遍历得到的序列是否是升序的
index = arrs[0];
boolean flag = true;
for(int i = 1; i < arr.size(); i++){
if(index >= arrs[i]){
flag = false;
break;
}
index = arrs[i];
}
return flag;
}

代码优化:

上述方法中,首先是将中序序列存储在集合中,为了判断升序方便又转换为数组,比较繁琐。

其实无需存储中序序列,在中序遍历过程中进行判断即可,一旦不符合升序条件就返回false。

代码:

    //代码优化(中序遍历过程中判断即可)
public static boolean isValidBST(TreeNode root){
if(root == null){
return true;
}
Stack<TreeNode> ts = new Stack<TreeNode>();
TreeNode ptr = root;
TreeNode pre = null;
//中序遍历
while(ptr != null || !ts.isEmpty()){
while(ptr != null){
ts.push(ptr);
ptr = ptr.left;
} if( !ts.isEmpty() ){
ptr = ts.pop();
if( pre != null && pre.val >= ptr.val){
return false;
}
pre = ptr;
ptr = ptr.right;
}
}
return true; }

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