LeetCode题解-110-Balanced Binary Tree

原题

原题链接：https://leetcode.com/problems/balanced-binary-tree/

这个题目有一个很大的争议就是AVL的定义，我以为的AVL的定义一直是“no 2 leaf nodes differ in distance from the root by more than 1”，与这个题目不同，所以一开始提交了错误的方法，但是查阅了资料发现LeetCode给出的定义应该才是正确的定义，请见https://discuss.leetcode.com/topic/276/two-different-definitions-of-balanced-binary-tree-result-in-two-different-judgments/2?show=63的讨论帖。

解法

解法分析

递归判断左子树是否是AVL，递归判断右子树是否为AVL，如果左右子树都是AVL，并且左右子树的高度差不超过1，那么这棵树为AVL。

代码

public class Solution110_recursive {    public boolean isBalanced(TreeNode root) {        return depth(root) != -1;    }        private int depth(TreeNode root){        if (root == null)            return 0;        int depthOfLeft = depth(root.left);        int depthOfRight = depth(root.right);        if (depthOfLeft == -1 || depthOfRight == -1)            return -1;        if (Math.abs(depthOfLeft - depthOfRight) > 1)            return -1;        return Math.max(depthOfLeft, depthOfRight) + 1;    }}

附

“no 2 leaf nodes differ in distance from the root by more than 1”的“AVL”的我的解法，未验证正确性。

解法分析

采用BFS，如果第i层未满并且第i+1层仍有节点，则非“avl”。

代码

public class Solution110_BFS_iterator {    public boolean isBalanced(TreeNode root) {        int expectSize = 1;        boolean levelNotFull = false;        Deque<TreeNode> deque = new LinkedList<TreeNode>();        if (root != null)            deque.push(root);        while (!deque.isEmpty()){            int exactSize = deque.size();            if (levelNotFull == true && exactSize != 0)                return false;            if (exactSize != expectSize)                levelNotFull = true;            expectSize = expectSize * 2;            while (exactSize-- > 0){                TreeNode currentNode = deque.pop();                if (currentNode.left != null)                    deque.addLast(currentNode.left);                if (currentNode.right != null)                    deque.addLast(currentNode.right);            }        }        return true;    }}