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[leetcode] 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.
https://oj.leetcode.com/problems/balanced-binary-tree/
思路：注意不要写成先序遍历的顺序，会造成多次遍历树，应该采用后续遍历的顺序依次判断。
注意java的没有引用传递，所以用了一个wrap类。
/* * be careful about the order of traversal!!!don‘t do recursion in recursion that resulting traverse the tree many times. * so we need to return the height and isBalanced of a node at the same time, but we can only return one, so if we use C++ * we can use reference, but it‘s java... so we have to wrap a primitive or using an array. * https://oj.leetcode.com/discuss/3931/can-we-have-a-better-solution * http://www.cnblogs.com/remlostime/archive/2012/10/27/2742987.html * */public class Solution { public boolean isBalanced(TreeNode root) { Height height = new Height(); return checkBalanced(root, height); } private boolean checkBalanced(TreeNode root, Height h) { if (root == null) { h.height = 0; return true; } Height lh = new Height(); Height rh = new Height(); boolean leftBalanced = checkBalanced(root.left, lh); boolean rightBalanced = checkBalanced(root.right, rh); h.height = Math.max(lh.height, rh.height) + 1; return leftBalanced && rightBalanced && (Math.abs(lh.height - rh.height) <= 1); } public static void main(String[] args) { TreeNode root = new TreeNode(5); root.right = new TreeNode(7); root.right.left = new TreeNode(3); root.right.left.right = new TreeNode(3); root.left = new TreeNode(3); System.out.println(new Solution().isBalanced(root)); } private static class Height { int height; }}
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