Invert Binary Tree 解答

Quetion

Invert a binary tree.

     4
   /   
  2     7
 /    / 
1   3 6   9

to

     4
   /   
  7     2
 /    / 
9   6 3   1

Solution 1 -- Recursion

Easy to think.

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

Solution 2 -- Iteration

BFS

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null)
            return root;
        List<TreeNode> current = new ArrayList<TreeNode>();
        List<TreeNode> next;
        current.add(root);
        while (current.size() > 0) {
            next = new ArrayList<TreeNode>();
            for (TreeNode tmpNode : current) {
                // inverse tmpNode
                TreeNode left = tmpNode.left;
                TreeNode right = tmpNode.right;
                tmpNode.right = left;
                tmpNode.left = right;
                if (right != null)
                    next.add(right);
                if (left != null)
                    next.add(left);
            }
            current = next;
        }
        return root;
    }
}