LeetCode(110) Balanced Binary Tree解题报告

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;
re
4000
turn 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;
}