LeetCode | Balanced Binary Tree

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

OJ's Binary Tree Serialization:    //LeetCode对二叉树的表示方法

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

1
/ \
2   3
/
4
\
5

The above binary tree is serialized as 

"{1,2,3,#,#,4,#,#,5}"

.

/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/

public class Solution {

public int maxDepth(TreeNode root){        //此方法为10.26日的题，递归求树的最大深度
if(root==null) return 0;
int leftDepth = maxDepth(root.left);
int rightDepth = maxDepth(root.right);
int depth =  leftDepth>=rightDepth?leftDepth:rightDepth ;
return depth+1;
}

public boolean isBalanced(TreeNode root) {
if (root==null) return true;               //注意，空树也认为是平衡树的情况

int leftHeight = maxDepth(root.left);       //求左右子树的高度
int rightHeight = maxDepth(root.right);

if( Math.abs(leftHeight-rightHeight)>1 ) return false;  //左右子树的高度相差大于1，一定不是平衡树
return isBalanced(root.left)&&isBalanced(root.right);   //注意！！！此处不能直接返回FALSE，要继续向下判断，
}                                                          //直至确认每个节点都满足平衡的条件才返回true

}
//左右子树的高度差不大于一，并不代表一定是平衡树，平衡树要求每个节点的左右子节点高度差都不大于1，不只是root节点
//例如：树{1 2 2 3 # # 3 4 # # 4}，其根的左右子树高度一样，但其不是平衡树(#代表此节点为null)