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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.

解题思路:

求左子树与右子树的深度之差，如果满足，分别递归左子树和右子树。但是这样有大量的重复计算，所以参考网上的一种做法，利用val来存深度，下面附上两次的代码。

public class Solution {    public boolean isBalanced(TreeNode root) {        if(root == null)            return true;        if(Math.abs(Depth(root.left)-Depth(root.right)) > 1)            return false;        return isBalanced(root.left) && isBalanced(root.right);      }    public int Depth(TreeNode root){        if(root == null)            return 0;        return Math.max(Depth(root.left)+1,Depth(root.right)+1);    }}

改进解法:

public class Solution {
       public boolean isBalanced(TreeNode root) {        if(root == null)            return true;        Depth(root);        return balanced(root);    }    public boolean balanced(TreeNode root){        int l=0,r=0;        if(root == null)            return true;        if(root.left != null) l = root.left.val;        if(root.right != null) r = root.right.val;        if(Math.abs(l-r) > 1)            return false;        return balanced(root.left) && balanced(root.right);    }    public int Depth(TreeNode root){        if(root == null)            return 0;        root.val = Math.max(Depth(root.left)+1,Depth(root.right)+1);        return root.val;    }

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