leetcode-tree

前序

class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        List<Integer> list = new LinkedList<>();
        if(root == null){
            return list;
        }
        //第一步是将根节点压入栈中
        stack.push(root);
        //当栈不为空时，出栈的元素插入list尾部。
        while(!stack.isEmpty()){
            root = stack.pop();
            list.add(root.val);
            if(root.right != null) stack.push(root.right);
            if(root.left != null) stack.push(root.left);
        }
        return list;
    }
}

中序

class Solution {
        public List<Integer> inorderTraversal(TreeNode root) {
            List<Integer> list = new ArrayList<>();
            Stack<TreeNode> stack = new Stack<>();
            TreeNode cur = root;
            while (cur != null || !stack.isEmpty()) {
                if (cur != null) {
                    stack.push(cur);
                    cur = cur.left;
                } else {
                    cur = stack.pop();
                    list.add(cur.val);
                    cur = cur.right;
                }
            }
            return list;
        }
}

后序

public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
    List<Integer> res = new ArrayList<Integer>();
    if(root == null)
        return res;
    Stack<TreeNode> stack = new Stack<TreeNode>();
    stack.push(root);
    while(!stack.isEmpty()){
        TreeNode node = stack.pop();
        if(node.left != null) stack.push(node.left);//和传统先序遍历不一样，先将左结点入栈
        if(node.right != null) stack.push(node.right);//后将右结点入栈
        res.add(0,node.val);                        //逆序添加结点值
    }     
    return res;
   }
}

层次

public List<List<Integer>> levelOrder(TreeNode root) {
    if(root == null)
        return new ArrayList<>();
    List<List<Integer>> res = new ArrayList<>();
    Queue<TreeNode> queue = new LinkedList<TreeNode>();
    queue.add(root);
    while(!queue.isEmpty()){
        int count = queue.size();
        List<Integer> list = new ArrayList<Integer>();
        while(count > 0){
            TreeNode node = queue.poll();
            list.add(node.val);
            if(node.left != null)
                queue.add(node.left);
            if(node.right != null)
                queue.add(node.right);
            count--;
        }
        res.add(list);
    }
    return res;
}

平衡二叉树

class Solution {
    private boolean isBalanced = true;
    public boolean isBalanced(TreeNode root) {
        getDepth(root);
        return isBalanced;
    }
    public int getDepth(TreeNode root) {
      if (root == null)
            return 0;
        int left = getDepth(root.left);
        int right = getDepth(root.right);
        if (Math.abs(left - right) > 1) {
            isBalanced = false;
        }
        return right > left ? right + 1 : left + 1;
      }
}

对称二叉树

class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        //把问题变成判断两棵树是否是对称的
        return isSym(root.left, root.right);
    }
    //判断的是根节点为r1和r2的两棵树是否是对称的
    public boolean isSym(TreeNode r1, TreeNode r2){
        if(r1 == null && r2 == null) return true;
        if(r1 == null || r2 == null) return false;
        //这两棵树是对称需要满足的条件：
        //1.俩根节点相等。 2.树1的左子树和树2的右子树，树2的左子树和树1的右子树都得是对称的
        return r1.val == r2.val && isSym(r1.left, r2.right) 
                            && isSym(r1.right, r2.left);
    }
}

重建二叉树

// 缓存中序遍历数组每个值对应的索引
private Map<Integer, Integer> indexForInOrders = new HashMap<>();

public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
    for (int i = 0; i < in.length; i++)
        indexForInOrders.put(in[i], i);
    return reConstructBinaryTree(pre, 0, pre.length - 1, 0);
}

private TreeNode reConstructBinaryTree(int[] pre, int preL, int preR, int inL) {
    if (preL > preR)
        return null;
    TreeNode root = new TreeNode(pre[preL]);
    int inIndex = indexForInOrders.get(root.val);
    int leftTreeSize = inIndex - inL;
    root.left = reConstructBinaryTree(pre, preL + 1, preL + leftTreeSize, inL);
    root.right = reConstructBinaryTree(pre, preL + leftTreeSize + 1, preR, inL + leftTreeSize + 1);
    return root;
}