LintCode : Balanced Binary Tree

Description:

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.

Example:

Given binary tree A={3,9,20,#,#,15,7}, B={3,#,20,15,7}

A)  3            B)    3 
   / \                  \
  9  20                 20
    /  \                / \
   15   7              15  7

The binary tree A is a height-balanced binary tree, but B is not.

/**
 * 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;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
    private boolean res = true;
    public boolean isBalanced(TreeNode root) {
        // write your code here
        if (root == null) {
            return true;
        }
        helper(root);
        if (res == false) {
            return false;
        }
        return true;
    }
    public void helper(TreeNode root) {
        if (root != null) {
            if (Math.abs(height(root.left) - height(root.right)) > 1) {
                res = false;
                return;
            }
            helper(root.left);
            helper(root.right);
        }
    }
    public int height(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return Math.max(height(root.left), height(root.right)) + 1;
    }
}

View Code

The trick here is to compare the height of the left child and right child.

转载于:https://www.cnblogs.com/dingjunnan/p/5408155.html