【LeetCode题解】二叉树的遍历

我准备开始一个新系列【LeetCode题解】，用来记录刷题，顺便复习一下数据结构与算法。

1. 二叉树

二叉树（binary tree）是一种极为普遍的数据结构，树的每一个节点最多只有两个节点——左孩子结点与右孩子结点。C实现的二叉树：

struct TreeNode {
	int val;
	struct TreeNode *left; // left child
	struct TreeNode *right; // right child
};

DFS

DFS的思想非常朴素：根据结点的连接关系，依次访问每一个节点，直至遍历完整棵树。根据根节点的访问次序的不同——前、中、后，可分为先序、中序、后序遍历。先序遍历是指先访问根节点，再依次访问左孩子节点、右孩子节点。下图给出一个二叉树的先序、中序、后序遍历示例：

递归实现：递归调用，打印根节点。C实现如下：

// preorder binary tree traversal
void preorder(struct TreeNode *root) {
	if(root) {
		printf("%d", root->val);
		preorder(root -> left);
		preorder(root -> right);
	}
}

void inorder(struct TreeNode *root) {
	if(root) {
		inorder(root -> left);
		printf("%d", root->val);
		inorder(root -> right);
	}
}

void postorder(struct TreeNode *root) {
	if(root) {
		postorder(root -> left);
		postorder(root -> right);
		printf("%d", root->val);
	}
}

任何递归实现的程序都可以改用栈实现。比如，先序遍历时用栈维护待访问结点；先将根结点入栈，再将右孩子结点入栈、左孩子结点入栈。Java实现如下：

public List<Integer> preorderTraversal(TreeNode root) {
  List<Integer> result = new ArrayList<>();
  if (root == null) return result;
  Stack<TreeNode> stack = new Stack<>();
  stack.push(root);
  while (!stack.isEmpty()) {
    TreeNode node = stack.pop();
    result.add(node.val);
    if (node.right != null) stack.push(node.right);
    if (node.left != null) stack.push(node.left);
  }
  return result;
}

中序遍历，栈维护已访问结点。一个结点只有当其左孩子结点已访问时，才能出栈，然后入栈右孩子结点；否则，则入栈左孩子结点。

public List<Integer> inorderTraversal(TreeNode root) {
  List<Integer> result = new ArrayList<>();
  if (root == null) return result;
  Stack<TreeNode> stack = new Stack<>();
  stack.push(root);
  while (!stack.isEmpty()) {
    TreeNode node = stack.peek();
    if (node.left != null) {
      stack.push(node.left);
      node.left = null; // its left child has been visited
    } else {
      result.add(node.val);
      stack.pop();
      if (node.right != null)
        stack.push(node.right);
    }
  }
  return result;
}

后序遍历，栈也维护已访问结点。一个结点只有当其左右孩子结点均已访问时，才能出栈；否则，则依次入栈右孩子结点、左孩子结点。

public List<Integer> postorderTraversal(TreeNode root) {
  List<Integer> result = new ArrayList<>();
  if (root == null) return result;
  Stack<TreeNode> stack = new Stack<>();
  stack.push(root);
  while (!stack.isEmpty()) {
    TreeNode node = stack.peek();
    if (node.left == null && node.right == null) {
      result.add(node.val);
      stack.pop();
    } else {
      if (node.right != null) {
        stack.push(node.right);
        node.right = null; // its right child has been visited
      }
      if (node.left != null) {
        stack.push(node.left);
        node.left = null; // its left child has been visited
      }
    }
  }
  return result;
}

层次遍历

层次遍历为二叉树的BFS，是指按照结点的层级依次从左至右访问。实现层序遍历需要借助于队列，用以维护待访问的结点。

public List<List<Integer>> levelOrder(TreeNode root) {
  List<List<Integer>> result = new LinkedList<>();
  if (root == null) return result;
  Queue<TreeNode> queue = new LinkedList<>();
  queue.offer(root);
  while (!queue.isEmpty()) {
    int levelSize = queue.size();
    List<Integer> levelList = new LinkedList<>();
    for (int i = 0; i < levelSize; i++) {
      TreeNode node = queue.poll();
      levelList.add(node.val);
      if (node.left != null) queue.offer(node.left);
      if (node.right != null) queue.offer(node.right);
    }
    result.add(levelList);
  }
  return result;
}

2. 题解

LeetCode题目	归类
100. Same Tree
101. Symmetric Tree
104. Maximum Depth of Binary Tree
111. Minimum Depth of Binary Tree
110. Balanced Binary Tree
112. Path Sum
113. Path Sum II
129. Sum Root to Leaf Numbers
144. Binary Tree Preorder Traversal	先序
94. Binary Tree Inorder Traversal	中序
145. Binary Tree Postorder Traversal	后序
102. Binary Tree Level Order Traversal	层次
107. Binary Tree Level Order Traversal II	层次
103. Binary Tree Zigzag Level Order Traversal	层次
114. Flatten Binary Tree to Linked List

100. Same Tree

判断两棵树是否一样，递归解决：

public boolean isSameTree(TreeNode p, TreeNode q) {
  if (p == null && q == null) return true;
  if (p == null || q == null) return false;
  return p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}

101. Symmetric Tree

同上一题较为类似，判断一棵树是否中心对称：

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

private boolean helper(TreeNode p, TreeNode q) {
  if (p == null && q == null) return true;
  if (p == null || q == null) return false;
  return p.val == q.val && helper(p.left, q.right) && helper(p.right, q.left);
}

104. Maximum Depth of Binary Tree

二叉树的最大深度，递归实现之：

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

111. Minimum Depth of Binary Tree

二叉树的最小深度，解决思路与上类似，不过有个special case——根不能视作为叶子节点：

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

110. Balanced Binary Tree

判断一棵树是否平衡，即每一个节点的左右子树的深度不大于1，正好可用到上面求树深度的代码：

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

112. Path Sum

判断root-to-leaf路径的结点之和是否存在，递归实现之：

public boolean hasPathSum(TreeNode root, int sum) {
  if (root == null) return false;
  if (root.val == sum && root.left == null && root.right == null) return true; // root is leaf node
  return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}

113. Path Sum II

上一个问题的变种，要将所有满足sum的情况保存下来；解决办法采用两个list存储中间结果，要注意list适当remove：

public List<List<Integer>> pathSum(TreeNode root, int sum) {
  List<List<Integer>> result = new LinkedList<>();
  List<Integer> pathList = new LinkedList<>(); // store a path which meets required sum
  helper(root, sum, result, pathList);
  return result;
}

private void helper(TreeNode root, int sum, List<List<Integer>> result, List<Integer> pathList) {
  if (root != null) {
    pathList.add(root.val);
    if (root.val == sum && root.left == null && root.right == null) {
      result.add(new LinkedList<>(pathList)); // deep copy
    } else {
      helper(root.left, sum - root.val, result, pathList);
      helper(root.right, sum - root.val, result, pathList);
    }
    pathList.remove(pathList.size() - 1); // remove the last element
  }
}

129. Sum Root to Leaf Numbers

二叉树的root-to-leaf路径表示一个十进制数，求所有路径所代表的数之和，递归实现：

public int sumNumbers(TreeNode root) {
  return helper(root, 0);
}

private int helper(TreeNode root, int sum) {
  if (root == null) return 0; // special case
  if (root.left == null && root.right == null) return sum * 10 + root.val;
  return helper(root.left, sum * 10 + root.val) + helper(root.right, sum * 10 + root.val);
}

144. Binary Tree Preorder Traversal

先序遍历，参看前面的栈实现。

94. Binary Tree Inorder Traversal

中序遍历，同上。

145. Binary Tree Postorder Traversal

后序遍历，同上。

102. Binary Tree Level Order Traversal

层次遍历，同上。

107. Binary Tree Level Order Traversal II

层次遍历变种，不同的是从bottom开始往上遍历；只需在层次遍历的代码中修改从头入result链表而不是从尾入：

result.addFirst(levelList);

103. Binary Tree Zigzag Level Order Traversa

“之”字形层次遍历，在层次遍历的基础上，将层级分为奇数、偶数，分别按顺序、逆序出队。

public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
  List<List<Integer>> result = new LinkedList<>();
  if (root == null) return result;
  Queue<TreeNode> queue = new LinkedList<>();
  queue.add(root);
  int level = 1;
  while (!queue.isEmpty()) {
    int levelSize = queue.size();
    List<Integer> levelList = new LinkedList<>();
    while (levelSize-- > 0) { // to deal with level nodes
      TreeNode node = queue.poll();
      levelList.add(node.val);
      if (node.left != null) queue.add(node.left);
      if (node.right != null) queue.add(node.right);
    }
    if (level % 2 == 0) // be stack order when level is even
      Collections.reverse(levelList);
    result.add(levelList);
    level++;
  }
  return result;
}

114. Flatten Binary Tree to Linked List

将二叉树打散成长链条树，从递归的角度来看，可分解为三个步骤：打散左子树，打散右子树，然后将右子树拼接到左子树尾部。

public void flatten(TreeNode root) {
  helper(root, null);
}

// flatten sub-tree, return its root node
private TreeNode helper(TreeNode root, TreeNode last) {
  if (root == null) return last;
  root.right = helper(root.left, helper(root.right, last));
  root.left = null;
  return root;
}