据结构与算法(八) 二叉树的练习

public int height() {
  return height(root);
}

private int  height(Node<E> node) {
  if (node == null) {
    return 0;
  }
  return 1 + Math.max(height(node.left), height(node.right));
}

public int height() {
        if (root == null) {
            return 0;
        }
        int height = 0;
        int  levelSize = 1;

        Queue<Node<E>> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            Node<E> node = queue.poll();
            levelSize--;

            if(node.left != null) {
                queue.offer(node.left);    
            }
            if(node.right!= null) {
                queue.offer(node.right);
            }    
            if (levelSize == 0) {
                levelSize = queue.size();
                height ++;
            }
        }
        return height;
    }

private boolean isCompleteTree(Node<E> root) {

  Queue<Node<E>> queue = new LinkedList<Node<E>>();
  queue.offer(root);
  boolean leaf = false;

  while (queue.size() != 0) {
    Node<E> node = queue.poll();

    if (leaf) {
      if (!node.leaf()) {
        return false;
      }
    }

    if (node.left != null) {
      queue.offer(node.left);
    }else {
      if (node.right != null) {
        return false;
      }
    }

    if (node.right != null) {
      queue.offer(node.right);
    }else {
      leaf = true;
    }
  }
  return true;
}

     4
   /   \
  2     7
 / \   / \
1   3 6   9

##### 输出：

     4
   /   \
  7     2
 / \   / \
9   6 3   1

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/invert-binary-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return root;
        }

        TreeNode tempNode = root.left;
        root.left = root.right;
        root.right = tempNode;

        invertTree(root.left);
        invertTree(root.right);

        return root;
    }
}