备战实习记录之【树与二叉树篇】4——翻转二叉树

4 226. 翻转二叉树

解法一：迭代实现前序遍历——栈

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        Stack<TreeNode> st = new Stack<>();
        if(root!=null) st.push(root);
        while(!st.empty()){
            TreeNode node = st.pop();
            TreeNode tmp = node.left;
            node.left = node.right;
            node.right = tmp;
            if(node.left!=null) st.push(node.left);
            if(node.right!=null) st.push(node.right);
        }
        return root;
    }
}

解法二：迭代统一写法 前序遍历——栈

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        Stack<TreeNode> st = new Stack<>();
        if(root!=null) st.push(root);
        while(!st.empty()){
            TreeNode node = st.peek();
            if(node!=null){
                st.pop();
                if(node.right!=null) st.push(node.right);
                if(node.left!=null) st.push(node.left);
                st.push(node);
                st.push(null);
            }else{
                st.pop();
                node = st.pop();
                TreeNode tmp = node.left;
                node.left = node.right;
                node.right = tmp;
            }
        }
        return root;
    }
}

解法三：递归实现

class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root==null) return root;
        TreeNode tmp = root.left;
        root.left = root.right;
        root.right = tmp;
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }
}