leetcode98. Validate Binary Search Tree

題目要求

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.

判斷一個樹是不是二叉查找樹。二叉查找樹即知足當前節點左子樹的值均小於當前節點的值，右子樹的值均大於當前節點的值。

思路一：stack dfs

能夠看到，對二叉查找樹的中序遍歷結果應當是一個遞增的數組。這裏咱們用堆棧的方式實現中序遍歷。

public boolean isValidBST(TreeNode root) {
        long currentVal = Long.MIN_VALUE;
        LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
        while(root!=null || !stack.isEmpty()){
            while(root!=null){
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            if(root.val<=currentVal) return false;
            currentVal = root.val;
            root = root.right;
        }
        return true;
    }

思路二：遞歸

咱們能夠發現，若是已知當前節點的值val以及取值的上下限upper，lower，那麼左子節點的取值範圍就是(lower, val),右子節點的取值範圍就是(val, upper)。由此出發遞歸判斷，時間複雜度爲O(n)由於每一個節點只須要遍歷一次。

public boolean isValidBST2(TreeNode root){
        return isValid(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    
    public boolean isValid(TreeNode treeNode, long lowerBound, long upperBound){
        if(treeNode==null) return true;
        if(treeNode.val>=upperBound || treeNode.val<=lowerBound) return false;
        return isValid(treeNode.left, lowerBound,treeNode.val) && isValid(treeNode.right, treeNode.val, upperBound);
    }

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