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;
    }
}
```