LeetCode Invert Binary Tree OJ 刷题 算法

Invert Binary Tree

Invert a binary tree.

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

Solution

import java.util.ArrayList;import java.util.List;public class InvertBinaryTree {    public static void main(String[] args) {        TreeNode r0 = new TreeNode(8);        TreeNode r11 = new TreeNode(4);        TreeNode r12 = new TreeNode(10);        TreeNode r21 = new TreeNode(2);        TreeNode r31 = new TreeNode(1);        TreeNode r23 = new TreeNode(9);        TreeNode r24 = new TreeNode(12);        r0.left = r11;        r0.right = r12;        r11.left = r21;        r12.left = r23;        r12.right = r24;        r21.left = r31;        // outputTree(r0);        // invertTree(r0);        // outputTree(r0);        //        TreeNode rr0 = new TreeNode(4);        TreeNode rr11 = new TreeNode(2);        TreeNode rr12 = new TreeNode(7);        TreeNode rr21 = new TreeNode(1);        TreeNode rr22 = new TreeNode(3);        TreeNode rr23 = new TreeNode(6);        TreeNode rr24 = new TreeNode(9);        rr0.left = rr11;        rr0.right = rr12;        rr11.left = rr21;        rr11.right = rr22;        rr12.left = rr23;        rr12.right = rr24;        // outputTree(rr0);        // invertTree(rr0);        // outputTree(rr0);        TreeNode q0 = new TreeNode(2);        TreeNode q11 = new TreeNode(3);        TreeNode q21 = new TreeNode(1);        q0.left = q11;        q11.left = q21;        outputTree(q0);        invertTree(q0);        outputTree(q0);    }    public static TreeNode invertTree(TreeNode root) {        if (root == null) {            return null;        }        if (root.left != null && root.right != null) {            int temp = root.left.val;            root.left.val = root.right.val;            root.right.val = temp;            TreeNode tempLeft = root.left.left;            root.left.left = root.right.left;            root.right.left = tempLeft;            TreeNode tempRight = root.left.right;            root.left.right = root.right.right;            root.right.right = tempRight;            invertTree(root.left);            invertTree(root.right);        }        if (root.left == null && root.right == null) {            return root;        }        if (root.left == null) {            root.left = new TreeNode(root.right.val);            root.left.left = root.right.left;            root.left.right = root.right.right;            root.right = null;            invertTree(root.left);            return root;        }        if (root.right == null) {            root.right = new TreeNode(root.left.val);            root.right.left = root.left.left;            root.right.right = root.left.right;            root.left = null;            invertTree(root.right);            return root;        }        return root;    }    public static void outputTree(TreeNode root) {        if (root == null) {            return;        }        System.out.println(root);        List<TreeNode> former = new ArrayList<TreeNode>();        former.add(root);        while (former.size() != 0) {            List<TreeNode> current = new ArrayList<TreeNode>();            for (TreeNode node : former) {                System.out.print(node.left + ", ");                System.out.print(node.right + ", ");                if (node.left != null) {                    current.add(node.left);                }                if (node.right != null) {                    current.add(node.right);                }            }            System.out.println();            former.clear();            former.addAll(current);        }    }}

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反思

这道题 错了几次, 自己做出来了

先是定义了一个输出 tree 的方法, 便于查看是否 invert 成功

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几次错误的原因

//wrong answerif (root.left == null) {        root.left = new TreeNode(root.right.val);        root.left.left = root.right.left;        root.left.right = root.right.right;        root.right = null;        //wrong answer    }if (root.right == null) {    root.right = new TreeNode(root.left.val);    root.right.left = root.left.left;    root.right.right = root.left.right;    root.left = null;    //wrong answer}

这儿是因为上面判断了 left 为 null 的情况, 但是将 right 置为null, 会进到下面的 right == null 的if 块 里

要在 if (root.left == null) {...} 最末加上
return root;

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第二次错误 原因

还是上面的代码块,

//wrong answerif (root.left == null) {        root.left = new TreeNode(root.right.val);        root.left.left = root.right.left;        root.left.right = root.right.right;        root.right = null;        //invertTree(root.left);        return root;    }if (root.right == null) {    root.right = new TreeNode(root.left.val);    root.right.left = root.left.left;    root.right.right = root.left.right;    root.left = null;    //invertTree(root.right);    return root;}

你会发现加了 return root 后, 如果 left 为空的话, 它和 right 交换后, 方法就返回了

它们的子节点并没有交换

要加上递归语句 invertTree(root.right);