[LeetCode]--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.

利用求最大深度，然后再求两边相减的绝对值如果是小于1的就行，然后如果所有子树都是平衡二叉树那么大树那必然是平衡二叉树。

/** * Definition for a binary tree node. * 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 (root.left == null && root.right != null && root.right.left == null && root.right.right == null) return true; if (root.left != null && root.right == null && root.left.left == null && root.left.right == null) return true; if (root.left != null && root.right != null && Math.abs(maxDepth(root.left) - maxDepth(root.right)) <= 1) return isBalanced(root.left) && isBalanced(root.right); else return false; } public int maxDepth(TreeNode root) { if (root == null) return 0; if (root.left == null && root.right == null) return 1; return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1; } }

别人写的代码第一个版本。

// Version 1: with ResultType /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ class ResultType { public boolean isBalanced; public int maxDepth; public ResultType(boolean isBalanced, int maxDepth) { this.isBalanced = isBalanced; this.maxDepth = maxDepth; } } public class Solution { /** * @param root: The root of binary tree. * @return: True if this Binary tree is Balanced, or false. */ public boolean isBalanced(TreeNode root) { return helper(root).isBalanced; } private ResultType helper(TreeNode root) { if (root == null) { return new ResultType(true, 0); } ResultType left = helper(root.left); ResultType right = helper(root.right); // subtree not balance if (!left.isBalanced || !right.isBalanced) { return new ResultType(false, -1); } // root not balance if (Math.abs(left.maxDepth - right.maxDepth) > 1) { return new ResultType(false, -1); } return new ResultType(true, Math.max(left.maxDepth, right.maxDepth) + 1); } }

我觉得第一个版本太啰嗦了，看第二个版本吧。这个真的是值得学习，这个if (left == -1 || right == -1 || Math.abs(left-right) > 1)判断语句值得深思。

// Version 2: without ResultType public class Solution { public boolean isBalanced(TreeNode root) { return maxDepth(root) != -1; } private int maxDepth(TreeNode root) { if (root == null) { return 0; } int left = maxDepth(root.left); int right = maxDepth(root.right); if (left == -1 || right == -1 || Math.abs(left-right) > 1) { return -1; } return Math.max(left, right) + 1; } }