二叉树 - Binary Tree

基本概念

• 根 (root)
• 叶子节点 (leaf)
• 子节点 (child)
• 节点的度 (degree)
• 树的高度 (height)
• 二叉树
  完全二叉树
  满二叉树
• 二叉树的性质
• 二叉搜索树 (BST)

设计与实现

• 节点

class TreeNode {
    int val;
    TreeNode left, right;
    TreeNode (int val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

• 二叉树遍历
  先序遍历

// version 1 : recursion

public class Solution {
    List<Integer> res;
    /**
     * @param root: A Tree
     * @return: Preorder in ArrayList which contains node values.
     */
    public List<Integer> preorderTraversal(TreeNode root) {
        res = new ArrayList<>();
        helper(root);
        return res;
    }
    private void helper(TreeNode root) {
        if (root == null) {
            return;
        }
        res.add(root.val);
        helper(root.left);
        helper(root.right);
    }
}

// version 2 : non - recursion (important!)

public class Solution {
    /**
     * @param root: A Tree
     * @return: Preorder in ArrayList which contains node values.
     */
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }        
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            res.add(cur.val);
            if (cur.right != null) {
                stack.push(cur.right);
            }
            if (cur.left != null) {
                stack.push(cur.left);
            }
        }
        return res;
    }
}

中序遍历

// version 1 : recursion

public class Solution {
    List<Integer> res;
    /**
     * @param root: A Tree
     * @return: Inorder in ArrayList which contains node values.
     */
    public List<Integer> inorderTraversal(TreeNode root) {
        res = new ArrayList<>();
        helper(root);
        return res;
    }
    private void helper(TreeNode root) {
        if (root == null) {
            return;
        }
        helper(root.left);
        res.add(root.val);
        helper(root.right);
    }
}

// version 2 : non - recursion (important!)

public class Solution {
    /**
     * @param root: A Tree
     * @return: Inorder in ArrayList which contains node values.
     */
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        while (root != null) {
            stack.push(root);
            root = root.left;
        }
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            res.add(cur.val);
            if (cur.right != null) {
                cur = cur.right;
                while (cur != null) {
                    stack.push(cur);
                    cur = cur.left;
                }
            }
        } 
        return res;
    }
}

后序遍历

// version 1 : recursion (by divide and conquer)

public class Solution {
    /**
     * @param root: A Tree
     * @return: Postorder in ArrayList which contains node values.
     */
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        res.addAll(postorderTraversal(root.left));
        res.addAll(postorderTraversal(root.right));
        res.add(root.val);
        return res;
    }
}

// version 2 : non - recursion (important!)

public class Solution {
    /**
     * @param root: A Tree
     * @return: Postorder in ArrayList which contains node values.
     */
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            res.add(cur.val);
            if (cur.left != null) {
                stack.push(cur.left);
            }
            if (cur.right != null) {
                stack.push(cur.right);
            }
        }
        Collections.reverse(res);
        return res;
    }
}

Morris遍历

// from: https://github.com/mission-peace/interview/blob/master/src/com/interview/tree/MorrisTraversal.java

public class MorrisTraversal {

    public void inorder(Node root) {
        Node current = root;
        while(current != null) {
            //left is null then print the node and go to right
            if (current.left == null) {
                System.out.print(current.data + " ");
                current = current.right;
            }
            else {
                //find the predecessor.
                Node predecessor = current.left;
                //To find predecessor keep going right till right node is not null or right node is not current.
                while(predecessor.right != current && predecessor.right != null){
                    predecessor = predecessor.right;
                }
                //if right node is null then go left after establishing link from predecessor to current.
                if(predecessor.right == null){
                    predecessor.right = current;
                    current = current.left;
                }else{ //left is already visit. Go rigth after visiting current.
                    predecessor.right = null;
                    System.out.print(current.data + " ");
                    current = current.right;
                }
            }
        }
    }

    public void preorder(Node root) {
        Node current = root;
        while (current != null) {
            if(current.left == null) {
                System.out.print(current.data + " ");
                current = current.right;
            }
            else {
                Node predecessor = current.left;
                while(predecessor.right != current && predecessor.right != null) {
                    predecessor = predecessor.right;
                }
                if(predecessor.right == null){
                    predecessor.right = current;
                    System.out.print(current.data + " ");
                    current = current.left;
                }else{
                    predecessor.right = null;
                    current = current.right;
                }
            }
        }
    }

}

Java中Balanced BST的实现

• TreeSet
  元素按照某种顺序被排序。
  基于HashMap实现，无容量限制。
  不允许元素重复。
  查找与删除的时间复杂度为。

Set<Integer> s = new TreeSet<>();

• TreeMap
  基于红黑树实现，无容量限制。

Lintcode 相关练习

• 基本题
  Maximum Depth of Binary Tree
  Binary Tree Leaf Sum
  Invert Binary Tree
  Binary Tree Path Sum
  Lowest Common Ancestor of a Binary Tree
• 二叉树上的分治算法
  Convert BST to Greater Tree
  Inorder Successor in BST
  Binary Tree Upside Down
  Find Leaves of Binary Tree
  Maximum Depth of Binary Tree