110.平衡二叉树

方法一：自顶向下递归

对于当前遍历到的节点，首先计算左右子树的高度，如果左右子树的高度差是否不超过 111，再分别递归地遍历左右子节点，并判断左子树和右子树是否平衡。这是一个自顶向下的递归的过程。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public boolean isBalanced(TreeNode root) {if(root == null){return true;}else{return Math.abs(height(root.left)- height(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right);}}public int height(TreeNode root){if(root == null){return 0;}else{return Math.max(height(root.left),height(root.right)) + 1;}}
}

方法二：自底向上递归

对于当前遍历到的节点，先递归地判断其左右子树是否平衡，再判断以当前节点为根的子树是否平衡。如果一棵子树是平衡的，则返回其高度（高度一定是非负整数），否则返回 −1。如果存在一棵子树不平衡，则整个二叉树一定不平衡。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public boolean isBalanced(TreeNode root) {return height(root) >= 0;}public int height(TreeNode root){if(root == null){return 0;}int left = height(root.left);int right = height(root.right);if(left == -1 || right == -1 || Math.abs(left - right) > 1){return -1;}else{return Math.max(left,right) + 1;}}
}