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android-notes/blogs/Algorithm/剑指 Offer/二叉树相关.md

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---
二叉树相关
---
#### 目录
1. #### [07. 重建二叉树](https://leetcode-cn.com/problems/zhong-jian-er-cha-shu-lcof/)
#### [07. 重建二叉树](https://leetcode-cn.com/problems/zhong-jian-er-cha-shu-lcof/)
```java
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
int n = preorder.length;
if (n == 0) {
return null;
}
int rootVal = preorder[0], rootIndex = 0;
for (int i = 0; i < n; i++) {
if (inorder[i] == rootVal) {
rootIndex = i;
break;
}
}
TreeNode root = new TreeNode(rootVal);
root.left = buildTree(Arrays.copyOfRange(preorder, 1, 1 + rootIndex),
Arrays.copyOfRange(inorder, 0, rootIndex));
root.right = buildTree(Arrays.copyOfRange(preorder, 1 + rootIndex, n),
Arrays.copyOfRange(inorder, rootIndex + 1, n));
return root;
}
}
```
```java
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
if (preorder == null || preorder.length == 0) {
return null;
}
TreeNode root = new TreeNode(preorder[0]);
int length = preorder.length;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
int inorderIndex = 0;
for (int i = 1; i < length; i++) {
int preorderVal = preorder[i];
TreeNode node = stack.peek();
if (node.val != inorder[inorderIndex]) {
node.left = new TreeNode(preorderVal);
stack.push(node.left);
} else {
while (!stack.isEmpty() && stack.peek().val == inorder[inorderIndex]) {
node = stack.pop();
inorderIndex++;
}
node.right = new TreeNode(preorderVal);
stack.push(node.right);
}
}
return root;
}
}
```