二叉树

本篇文章记录常见的二叉树算法题

leetcode94 二叉树的中序遍历

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while(!stack.isEmpty() || cur!=null){
            while(cur!=null){
                stack.push(cur);
                cur=cur.left;
            }
            cur = stack.pop();
            res.add(cur.val);
            cur = cur.right;

        }

        return res;

    }
}

这是二叉树的最经典问题之一，BST的相关问题可以沿用该思路

leetcode104 二叉树的最大深度

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int maxDepth(TreeNode root) {
        if(root == null){
            return 0;
        }

        int left = maxDepth(root.left);
        int right = maxDepth(root.right);
        return Math.max(left,right)+1;
    }

}

二叉树大部分题目可以考虑使用递归，这题是一个简单案例

leetcode236 二叉树的最近公共祖先

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null || root == p || root == q){
            return root;
        }

        TreeNode left = lowestCommonAncestor(root.left,p,q);
        TreeNode right = lowestCommonAncestor(root.right,p,q);
        if(left!=null &&right!=null){
            return root;
        }
        if(left!=null){
            return left;
        }

        return right;
    }
}

这个题也是递归的使用

leetcode102 二叉树的层序遍历

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        if(root == null){
            return res;
        }

        queue.offer(root);

        while(!queue.isEmpty()){
            List<Integer> temp = new ArrayList<>();
            int size = queue.size();
            /** 表示一层 */
            for(int i = 0;i<size;i++){
                TreeNode node = queue.poll();
                temp.add(node.val);

                if(node.left!=null){
                    queue.offer(node.left);
                }
                if(node.right!=null){
                    queue.offer(node.right);
                }   
            }
            res.add(temp);

        }
        return res;
    }
}

这题目也是比较经典的题，类似求树的深度，右视图等都可以使用该思路

leetcode105 从前序与中序遍历序列构造二叉树

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder.length == 0||inorder.length == 0){
            return null;
        }
        TreeNode root = new TreeNode(preorder[0]);
        for(int i=0;i<preorder.length;i++){
            if(inorder[i] == preorder[0]){
                root.left = buildTree(Arrays.copyOfRange(preorder,1,1+i),Arrays.copyOfRange(inorder,0,i));
                root.right = buildTree(Arrays.copyOfRange(preorder,1+i,preorder.length),Arrays.copyOfRange(inorder,i+1,inorder.length));
                break;
            }
        }

        return root;
    }
}