To prove safety offer 39. balanced binary tree

To prove safety offer 39. balanced binary tree

topic

Input binary tree, the binary tree is determined whether a balanced binary tree.

Thinking

Lazy continued, with a code of a preceding question, seeking depth, compare left and right subtrees depth before returning, if the difference is greater than 1, the result is set to false.

Code

  public class TreeNode {

    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
      this.val = val;

    }

  }

  boolean ans = true;

  public int TreeDepth(TreeNode root) {
    if (root == null) {
      return 0;
    }
    int left = TreeDepth(root.left);
    int right = TreeDepth(root.right);
    if (Math.abs(left - right) > 1) {
      ans = false;
    }
    return Math.max(left, right) + 1;
  }

  public boolean IsBalanced_Solution(TreeNode root) {
    int i = TreeDepth(root);
    return ans;
  }