Invert Binary Tree

ID: 226; easy

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LintCode 炼码

Solution 1 (Go)

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func invertTree(root *TreeNode) *TreeNode {
    if root == nil {
        return nil
    }
    invertTree(root.Left)
    invertTree(root.Right)
    root.Left, root.Right = root.Right, root.Left
    return root
}

Solution 2 (Java)

/**
 * 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: a TreeNode, the root of the binary tree
     * @return: nothing
     */
    public void invertBinaryTree(TreeNode root) {
        if (root == null) return;

        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;

        invertBinaryTree(root.left);
        invertBinaryTree(root.right);
    }
}

Notes

• Recursion

Solution 3 (Java)

/**
 * 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: a TreeNode, the root of the binary tree
     * @return: nothing
     */
    public void invertBinaryTree(TreeNode root) {
        if (root == null) return;
        Queue<TreeNode> q = new ArrayDeque<TreeNode>();
        q.offer(root);

        while (!q.isEmpty()) {
            TreeNode curr = q.poll();
            TreeNode temp = curr.left;
            curr.left = curr.right;
            curr.right = temp;

            if (curr.left != null)
                q.offer(curr.left);
            if (curr.right != null)
                q.offer(curr.right);
        }
    }
}

Notes

• BFS