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

---
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/)
2. [26. 树的子结构](https://leetcode-cn.com/problems/shu-de-zi-jie-gou-lcof/)
3. [ 27. 二叉树的镜像](https://leetcode-cn.com/problems/er-cha-shu-de-jing-xiang-lcof/)
4. [32 - I. 从上到下打印二叉树](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-lcof/)
5. [ Offer 34. 二叉树中和为某一值的路径](https://leetcode-cn.com/problems/er-cha-shu-zhong-he-wei-mou-yi-zhi-de-lu-jing-lcof/)
6. [55 - I. 二叉树的深度](https://leetcode-cn.com/problems/er-cha-shu-de-shen-du-lcof/)
7. [55 - II. 平衡二叉树](https://leetcode-cn.com/problems/ping-heng-er-cha-shu-lcof/)
8. [32 - II. 从上到下打印二叉树 II](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-ii-lcof/)
9. [32 - III. 从上到下打印二叉树 III](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-iii-lcof/)
10. [37. 序列化二叉树](https://leetcode-cn.com/problems/xu-lie-hua-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;
    }
}
```

#### [32 - I. 从上到下打印二叉树](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-lcof/)

```java
class Solution {

    public int[] levelOrder(TreeNode root) {
        if (root == null) {
            return new int[0];
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        List<Integer> list = new ArrayList<>();
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            list.add(node.val);
            if (node.left != null) {
                queue.add(node.left);
            }
            if (node.right != null) {
                queue.add(node.right);
            }
        }
        int[] result = new int[list.size()];
        int index = 0;
        for (int i : list) {
            result[index++] = i;
        }
        return result;
    }
}
```

#### [ Offer 34. 二叉树中和为某一值的路径](https://leetcode-cn.com/problems/er-cha-shu-zhong-he-wei-mou-yi-zhi-de-lu-jing-lcof/)

```java
class Solution {

    List<List<Integer>> result = new ArrayList<>();
    List<Integer> path = new ArrayList<>();

    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        path(root, sum);
        return result;
    }

    private void path(TreeNode root, int sum) {
        if (root == null) {
            return;
        }
        path.add(root.val);
        int target = sum - root.val;
        if (target == 0 && root.left == null && root.right == null) {
            result.add(new ArrayList(path));
        }
        path(root.left, target);
        path(root.right, target);
        path.remove(path.size() - 1);
    }
}
```

#### [55 - I. 二叉树的深度](https://leetcode-cn.com/problems/er-cha-shu-de-shen-du-lcof/)

```java
class Solution {

    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
    }
}
```

```java
class Solution {

    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int result = 0;
        Deque<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            result++;
            int n = queue.size();
            for (int i = 0; i < n; i++) {
                TreeNode node = queue.poll();
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
            }
        }
        return result;
    }
}
```

#### [55 - II. 平衡二叉树](https://leetcode-cn.com/problems/ping-heng-er-cha-shu-lcof/)

```java
class Solution {

    public boolean isBalanced(TreeNode root) {
        if (root == null) {
            return true;
        }
        Deque<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            int offset = Math.abs(maxDepth(node.left) - maxDepth(node.right));
            if (offset > 1) {
                return false;
            }
            if (node.left != null) {
                queue.add(node.left);
            }
            if (node.right != null) {
                queue.add(node.right);
            }
        }
        return true;
    }

    private int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
    }
}
```

```java
class Solution {

    public boolean isBalanced(TreeNode root) {
        if (root == null) {
            return true;
        }
        return Math.abs(maxDepth(root.left) - maxDepth(root.right)) <= 1 && isBalanced(root.left) && isBalanced(
                root.right);
    }

    private int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
    }
}
```

#### [32 - II. 从上到下打印二叉树 II](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-ii-lcof/)


```java
class Solution {

    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Deque<TreeNode> deque = new LinkedList<>();
        deque.add(root);
        while (!deque.isEmpty()) {
            int l = deque.size();
            List<Integer> list = new ArrayList<>();
            for (int i = 0; i < l; i++) {
                TreeNode node = deque.poll();
                list.add(node.val);
                if (node.left != null) {
                    deque.add(node.left);
                }
                if (node.right != null) {
                    deque.add(node.right);
                }
            }
            result.add(list);
        }
        return result;
    }
}
```

#### [32 - III. 从上到下打印二叉树 III](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-iii-lcof/)

```java
class Solution {

    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        Deque<TreeNode> deque = new LinkedList<>();
        deque.add(root);
        while (!deque.isEmpty()) {
            int l = deque.size();
            List<Integer> list = new ArrayList<>(l);
            for (int i = 0; i < l; i++) {
                TreeNode node = deque.poll();
                list.add(node.val);
                if (node.left != null) {
                    deque.add(node.left);
                }
                if (node.right != null) {
                    deque.add(node.right);
                }
            }
            if (res.size() % 2 == 1) {
                Collections.reverse(list);
            }
            res.add(list);
        }
        return res;
    }
}
```

#### [37. 序列化二叉树](https://leetcode-cn.com/problems/xu-lie-hua-er-cha-shu-lcof/)

```java
public class Codec {

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        if (root == null) {
            return "[]";
        }
        Deque<TreeNode> deque = new LinkedList<>();
        deque.add(root);
        StringBuilder builder = new StringBuilder("[");
        while (!deque.isEmpty()) {
            TreeNode node = deque.poll();
            if (node != null) {
                builder.append(node.val).append(",");
                deque.add(node.left);
                deque.add(node.right);
            } else {
                builder.append("null,");
            }
        }
        builder.deleteCharAt(builder.length() - 1);
        builder.append("]");
        return builder.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if ("[]".equals(data)) {
            return null;
        }
        String[] vals = data.substring(1, data.length() - 1).split(",");
        TreeNode root = new TreeNode(Integer.parseInt(vals[0]));
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        int i = 1;
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if (!vals[i].equals("null")) {
                node.left = new TreeNode(Integer.parseInt(vals[i]));
                queue.add(node.left);
            }
            i++;
            if (!vals[i].equals("null")) {
                node.right = new TreeNode(Integer.parseInt(vals[i]));
                queue.add(node.right);
            }
            i++;
        }
        return root;
    }
}
```

#### [28. 对称的二叉树](https://leetcode-cn.com/problems/dui-cheng-de-er-cha-shu-lcof/)

```java
class Solution {

    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        return helper(root.left, root.right);
    }

    private boolean helper(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 == null) {
            return true;
        }
        if (root1 == null || root2 == null) {
            return false;
        }
        if (root1.val != root2.val) {
            return false;
        }
        return helper(root1.left, root2.right) && helper(root1.right, root2.left);
    }
}
```

```java
class Solution {

    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        Deque<TreeNode> queue = new LinkedList<>();
        queue.addFirst(root.left);
        queue.addLast(root.right);
        while (!queue.isEmpty()) {
            TreeNode first = queue.removeFirst();
            TreeNode last = queue.removeLast();
            if (first == null && last == null) {
                continue;
            }
            if (first == null || last == null) {
                return false;
            }
            if (first.val != last.val) {
                return false;
            }
            queue.addFirst(first.right);
            queue.addFirst(first.left);
            queue.addLast(last.left);
            queue.addLast(last.right);
        }
        return true;
    }
}
```

#### [33. 二叉搜索树的后序遍历序列](https://leetcode-cn.com/problems/er-cha-sou-suo-shu-de-hou-xu-bian-li-xu-lie-lcof/)

```java
class Solution {

    public boolean verifyPostorder(int[] postorder) {
        return helper(postorder, 0, postorder.length - 1);
    }

    private boolean helper(int[] order, int i, int j) {
        if (i >= j) {
            return true;
        }
        int p = i;
        while (order[p] < order[j]) {
            p++;
        }
        int m = p;
        while (order[p] > order[j]) {
            p++;
        }
        return p == j && helper(order, i, m - 1) && helper(order, m, j - 1);
    }
}
```