原题链接：

110. 平衡二叉树

https://leetcode.cn/problems/balanced-binary-tree/description/

完成情况：

解题思路：

就是要算高度，然后比较每一次迭代过程中的高度。

参考代码：

递归法：

package 代码随想录.树.二叉树;import 代码随想录.树.TreeNode;import java.util.LinkedList;
import java.util.Map;
import java.util.Queue;public class _110平衡二叉树_递归法 {/***一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1 。** @param root* @return*/public boolean isBalanced(TreeNode root) {return getHeight(root)!= -1;}/**** @param root* @return*/private int getHeight(TreeNode root) {if (root == null) {return 0;}int leftHeight = getHeight(root.left);if (leftHeight == -1) {return -1;}int rightHeight = getHeight(root.right);if (rightHeight == -1) {return -1;}if (Math.abs(leftHeight - rightHeight) > 1) {return -1;}return Math.max(leftHeight, rightHeight) + 1;}
}

迭代法

package 代码随想录.树.二叉树;import 代码随想录.树.TreeNode;import java.util.ArrayDeque;
import java.util.Deque;
import java.util.LinkedList;
import java.util.Queue;public class _110平衡二叉树_迭代 {/**** @param root* @return*/public boolean isBalanced(TreeNode root) {if (root == null) {return true;}Deque<TreeNode> myStack = new LinkedList<TreeNode>();TreeNode prev = null;while (!myStack.isEmpty() || root!= null) {while (root!= null) {myStack.push(root);root = root.left;}TreeNode node = myStack.peek();// 右结点为null或已经遍历过if (node.right == null || node.right == prev) {if (Math.abs(getHeight(node.left) - getHeight(node.right)) > 1) {return false;}myStack.pop();prev = node;root = null;// 当前结点下，没有要遍历的结点了}else {root = node.right;}}return true;}/*** 层序遍历，求结点的高度** @param node* @return*/private int getHeight(TreeNode node) {if (node == null) {return 0;}Deque<TreeNode> queue = new LinkedList<TreeNode>();queue.offer(node);int height = 0;while (!queue.isEmpty()) {int size = queue.size();height++;while (size > 0) {TreeNode curNode = queue.poll();if (curNode.left!= null) {queue.offer(curNode.left);}if (curNode.right!= null) {queue.offer(curNode.right);}}}return height;}
}

错误经验吸取