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.
Solution 1:
//相当于中序遍历,先加入中间节点,再加入左节点,再加入右节点
class Solution {
    public boolean isValidBST(TreeNode root) {
        if (root == null)
            return true;
        Stack<TreeNode> s = new Stack<>();
        TreeNode pre = null;
        s.push(root);
        while (!s.isEmpty()) {
            TreeNode tn = s.peek();
            while (tn != null) {
                tn = tn.left;                
                s.push(tn);                            
            }
            s.pop();
            if (!s.isEmpty()) {
                tn = s.pop();
                if (pre != null && tn.val <= pre.val)
                   return false;
                pre = tn;            
                s.push(tn.right);
            }            
        }
        return true;
    }
}
 
Solution 2://必须是Long类型,例如只有有个2^31-1节点,如果把Long改成Integer,则会报错
class Solution {
    Integer prev = null;
    boolean isValid = true;
    public boolean isValidBST(TreeNode root) {
        helper(root);
        return isValid;
    }
    
    private void helper(TreeNode root) {
        if (root == null)
            return;
        if (isValid == false)
            return;
        helper(root.left);
        if (prev != null) {
            if (root.val <= prev)
                isValid = false;
        }
        prev = root.val;
        helper(root.right);               
    }
}

原文地址:http://www.cnblogs.com/MarkLeeBYR/p/16906689.html

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