LeetCode OJ 110. 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.
做完了链表系列,这几天开始做树系列。对于解决关于树的问题,递归是一个很好的方法。对于这个题目我们也可以用递归来解决,给定一个节点,我们分别判断它的左子树是否平衡,右子树是否平衡。如果它的子树有不平衡的,那么该树是不平衡的。如果它的子树平衡,我们要比较一下两个子树深度差是否超过1。如果超过1,那么该树也是不平衡的。如何记录节点的深度呢,我们可以利用节点中的val来存储一个节点和它的子树组成的树的深度。是不是很巧妙呢?代码如下:
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public boolean isBalanced(TreeNode root) { 12 if(root == null) return true; 13 if(root.left == null && root.right == null){ 14 root.val = 1; 15 return true; 16 } 17 18 int leftdeep = 0; 19 int rightdeep = 0; 20 if(root.left != null){ 21 if(!isBalanced(root.left)) return false; 22 leftdeep = root.left.val; 23 } 24 if(root.right != null){ 25 if(!isBalanced(root.right)) return false; 26 rightdeep = root.right.val; 27 } 28 29 if(Math.abs(leftdeep - rightdeep) > 1) return false; 30 root.val = leftdeep>rightdeep?leftdeep+1:rightdeep+1; 31 return true; 32 } 33 }