algorithm/leetcode-master
1
0
Fork 0
mirror of https://github.com/youngyangyang04/leetcode-master.git synced 2025-07-07 07:35:35 +08:00
Code Issues Packages Projects Releases Wiki Activity

Merge pull request #1505 from w2xi/master

Browse Source 
Update 0226.翻转二叉树.md JavaScript递归版本
This commit is contained in:
程序员Carl
2022-07-13 09:44:33 +08:00
committed by GitHub
parent e3a0207d02 0ee18c392d
commit b8c3d4541c
2 changed files with 41 additions and 41 deletions
Download Patch File Download Diff File Expand all files Collapse all files

25
problems/0226.翻转二叉树.md
View File

@ -470,25 +470,14 @@ func invertTree(root *TreeNode) *TreeNode {
使用递归版本的前序遍历
```javascript
var invertTree = function(root) {
//1. 首先使用递归版本的前序遍历实现二叉树翻转
//交换节点函数
const inverNode=function(left,right){
let temp=left;
left=right;
right=temp;
//需要重新给root赋值一下
root.left=left;
root.right=right;
// 终止条件
if (!root) {
return null;
}
//确定递归函数的参数和返回值inverTree=function(root)
//确定终止条件
if(root===null){
return root;
}
//确定节点处理逻辑 交换
inverNode(root.left,root.right);
invertTree(root.left);
invertTree(root.right);
// 交换左右节点
const rightNode = root.right;
root.right = invertTree(root.left);
root.left = invertTree(rightNode);
return root;
};
```

57
problems/0450.删除二叉搜索树中的节点.md
View File

@ -456,31 +456,42 @@ func deleteNode(root *TreeNode, key int) *TreeNode {
* @param {number} key
* @return {TreeNode}
*/
var deleteNode = function (root, key) {
if (root === null)
return root;
if (root.val === key) {
if (!root.left)
return root.right;
else if (!root.right)
return root.left;
else {
let cur = root.right;
while (cur.left) {
cur = cur.left;
}
cur.left = root.left;
root = root.right;
delete root;
return root;
}
}
if (root.val > key)
root.left = deleteNode(root.left, key);
if (root.val < key)
var deleteNode = function(root, key) {
if (!root) return null;
if (key > root.val) {
root.right = deleteNode(root.right, key);
return root;
return root;
} else if (key < root.val) {
root.left = deleteNode(root.left, key);
return root;
} else {
// 场景1: 该节点是叶节点
if (!root.left && !root.right) {
return null
}
// 场景2: 有一个孩子节点不存在
if (root.left && !root.right) {
return root.left;
} else if (root.right && !root.left) {
return root.right;
}
// 场景3: 左右节点都存在
const rightNode = root.right;
// 获取最小值节点
const minNode = getMinNode(rightNode);
// 将待删除节点的值替换为最小值节点值
root.val = minNode.val;
// 删除最小值节点
root.right = deleteNode(root.right, minNode.val);
return root;
}
};
function getMinNode(root) {
while (root.left) {
root = root.left;
}
return root;
}
```
迭代