失败：Flatten Binary Tree to Linked List

来源：互联网发布：mac只装win10单系统编辑：程序博客网时间：2024/06/10 21:10

问题：

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

         1        / \       2   5      / \   \     3   4   6

The flattened tree should look like:

   1    \     2      \       3        \         4          \           5            \             6

12分钟出来的暴力解法，但是导致了Time Limit Exceed

/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public void flatten(TreeNode root) {
if (root == null || isLeaf(root)) {
return;
}

root.right = connectFlat(flat(root.left), flat(root.right));
root.left = null;
}

private boolean isLeaf(TreeNode node) {
if (node == null) {
return false;
}

return node.left == null && node.right == null;
}
private TreeNode flat(TreeNode node) {
if (node == null) {
} else if (isLeaf(node)) {
} else if (isLeaf(node.left)) {
TreeNode originRight = node.right;
node.right = node.left;
node.left = null;
node.right.right = flat(originRight);
} else if (isLeaf(node.right)) {
node.right.right = flat(node.left);
} else {
node.right = connectFlat(flat(node.left), flat(node.right));
}

return node;
}

private TreeNode connectFlat(TreeNode node1, TreeNode node2) {
TreeNode currentNode = node1;

if (node1 == null) {
return node2;
}

while (currentNode.right != null) {
currentNode = currentNode.right;
}

currentNode.right = node2;
return node1;
}
}

20分钟 again

/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private TreeNode currentNode;
public void flatten(TreeNode root) {
if (root == null) {
return;
}

if (root.left == null && root.right == null) {
return;
}
TreeNode currentNode = root;
TreeNode currentAddNode = root;

while (currentNode != null) {
if (currentNode.left != null) {
TreeNode tempRight = currentNode.right;
currentNode.right = currentNode.left;
findAddNode(currentNode.left).right = tempRight;
currentNode.left = null;
}
currentNode = currentNode.right;
}
}

private TreeNode findAddNode(TreeNode node) {
if (node == null) {
return null;
}
while (node.right != null) {
node = node.right;
}

return node;
}
}

0 0