【LeetCode】Balanced Binary Tree 解题报告
2014-10-12 20:15
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【题目】
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.
【说明】
不是很难,思路大家可能都会想到用递归,分别判断左右两棵子树是不是平衡二叉树,如果都是并且左右两颗子树的高度相差不超过1,那么这棵树就是平衡二叉树。
【比较直观的Java代码】
【改进后】
利用了TreeNode结构中的val,用它来记录以当前结点为根的子树的高度,避免多次计算。
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.
【说明】
不是很难,思路大家可能都会想到用递归,分别判断左右两棵子树是不是平衡二叉树,如果都是并且左右两颗子树的高度相差不超过1,那么这棵树就是平衡二叉树。
【比较直观的Java代码】
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public boolean isBalanced(TreeNode root) { if(root == null){ return true; } if(root.left==null && root.right==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 1+Math.max(depth(root.left), depth(root.right)); } }
【改进后】
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public boolean isBalanced(TreeNode root) { height(root); return run(root); } public boolean run(TreeNode root) { if (root == null) return true; int l = 0, r = 0; if (root.left != null) l = root.left.val; if (root.right != null) r = root.right.val; if (Math.abs(l - r) <= 1 && isBalanced(root.left) && isBalanced(root.right)) return true; return false; } public int height(TreeNode root) { if (root == null) return 0; root.val = Math.max( height(root.left), height(root.right) ) + 1; return root.val; } }
利用了TreeNode结构中的val,用它来记录以当前结点为根的子树的高度,避免多次计算。
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