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;
    }
}
```

#### [26. 树的子结构](https://leetcode-cn.com/problems/shu-de-zi-jie-gou-lcof/)

```java
class Solution {

    public boolean isSubStructure(TreeNode A, TreeNode B) {
        if (A == null || B == null) {
            return false;
        }
        return isEquals(A, B) || isSubStructure(A.left, B) || isSubStructure(A.right, B);
    }

    private boolean isEquals(TreeNode nodeA, TreeNode nodeB) {
        if (nodeB == null) {
            return true;
        }
        if (nodeA == null || nodeA.val != nodeB.val) {
            return false;
        }
        return isEquals(nodeA.left, nodeB.left) && isEquals(nodeA.right, nodeB.right);
    }
}
```

#### [ 27. 二叉树的镜像](https://leetcode-cn.com/problems/er-cha-shu-de-jing-xiang-lcof/)

```java
class Solution {

    public TreeNode mirrorTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        TreeNode rootLeft = mirrorTree(root.right);
        TreeNode rootRight = mirrorTree(root.left);
        root.left = rootLeft;
        root.right = rootRight;
        return root;
    }
}
```

```java
class Solution {

    public TreeNode mirrorTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        Stack<TreeNode> stack = new Stack<>();
        stack.add(root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
			if (node.left != null) {
				stack.add(node.left);
			}
			if (node.right != null) {
				stack.add(node.right);
			}
            TreeNode temp = node.left;
            node.left = node.right;
            node.right = temp;
        }
        return root;
    }
}
```