代码随想录算法训练营第十八天| 513. 找树左下角的值 112. 路径总和 113. 路径总和 II 106. 从中序与后序遍历序列构造二叉树 105. 从前序与中序遍历序列构造二叉树

513. 找树左下角的值
https://leetcode.cn/problems/find-bottom-left-tree-value/description/

public int findBottomLeftValue(TreeNode root) {
        int val = 0;
        Deque<TreeNode> deque = new ArrayDeque<>();
        deque.offer(root);
        while (!deque.isEmpty()){
            int len = deque.size();
            for (int i = 0; i < len; i++) {
                TreeNode node = deque.poll();
                if (i == 0) val = node.val;
                if (node.left != null) deque.offer(node.left);
                if (node.right != null) deque.offer(node.right);
            }
        }
        return val;
    }

总结：层序遍历
112. 路径总和
https://leetcode.cn/problems/path-sum/description/

public boolean hasPathSum(TreeNode root, int targetSum) {
        if (root == null) return false;
        boolean sum = sum(root, targetSum);
        return sum;
    }
    public boolean sum(TreeNode node,int sum){
        if (node.left == null && node.right == null) {
            return sum - node.val == 0; // 到达叶子节点时判断路径和是否等于目标和
        }
        boolean sumLeft = false;
        boolean sumRight = false;
        if (node.left != null) {
            sumLeft = sum(node.left, sum - node.val);
        }
        if (node.right != null){
            sumRight = sum(node.right, sum - node.val);
        }
        return sumLeft || sumRight;
    }

总结：递归到叶子节点，判断是否合理
113. 路径总和 II
https://leetcode.cn/problems/path-sum-ii/description/

public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        List<List<Integer>> resList = new ArrayList<>();
        List<Integer> path = new ArrayList<>();
        if (root == null) return resList;
        pathSSum(root,targetSum,resList,path);
        return resList;
    }
    public void pathSSum(TreeNode node,int targetSum,List<List<Integer>> resList,List<Integer> path){
        if (node.left == null && node.right == null){
            if (targetSum - node.val == 0){
                path.add(node.val);
                resList.add(new ArrayList<>(path));
                path.remove(path.size() - 1); //回溯
            }
            return;
        }
        if (node.left != null){
            path.add(node.val);
            pathSSum(node.left,targetSum - node.val,resList,path);
            path.remove(path.size() - 1); //回溯
        }
        if (node.right != null){
            path.add(node.val);
            pathSSum(node.right,targetSum - node.val,resList,path);
            path.remove(path.size() - 1); //回溯
        }
    }

总结：带了点回溯
106. 从中序与后序遍历序列构造二叉树
https://leetcode.cn/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/

int post_idx;
    int[] postorder;
    int[] inorder;
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();

    public TreeNode buildTree(int[] inorder, int[] postorder) {
        this.postorder = postorder;
        this.inorder = inorder;
        // 从后序遍历的最后一个元素开始
        post_idx = postorder.length - 1;

        // 建立（元素，下标）键值对的哈希表
        for (int i = 0; i < inorder.length; i++) {
            map.put(inorder[i],i);
        }

        return helper(0, inorder.length - 1);
    }
    public TreeNode helper(int in_left, int in_right) {
        // 如果这里没有节点构造二叉树了，就结束
        if (in_left > in_right) {
            return null;
        }

        // 选择 post_idx 位置的元素作为当前子树根节点
        int root_val = postorder[post_idx];
        TreeNode root = new TreeNode(root_val);

        // 根据 root 所在位置分成左右两棵子树
        int index = map.get(root_val);

        // 下标减一
        post_idx--;
        // 构造右子树
        root.right = helper(index + 1, in_right);
        // 构造左子树
        root.left = helper(in_left, index - 1);
        return root;
    }

总结：力扣官方的代码，递归中每次传中序数组的左右边界，每次递归把当前的root拿出来去给他构造left，right。返回当前root，注意这里要先构造右子树，再左子树，因为后序数组的遍历是先右再左的。
105. 从前序与中序遍历序列构造二叉树
https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/

int[] preorder;
    int[] inorder;
    int pre_index;
    HashMap<Integer,Integer> map;
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        this.preorder = preorder;
        this.inorder = inorder;
        map = new HashMap<>();
        for (int i = 0; i < inorder.length; i++) {
            map.put(inorder[i],i);
        }
        pre_index = 0;
        return helper(0,inorder.length - 1);
    }
    public TreeNode helper(int left,int right){
        if (left > right) return null;
        int concurrentRootVal = preorder[pre_index];
        pre_index++;
        TreeNode node = new TreeNode(concurrentRootVal);
        int index = map.get(concurrentRootVal);
        node.left = helper(left,index - 1);
        node.right = helper(index + 1 ,right);
        return node;
    }

总结：力扣官方的代码，递归中每次传中序数组的左右边界，每次递归把当前的root拿出来去给他构造left，right。返回当前root，注意这里要先构造左子树，再右子树，因为前序数组的遍历是先左再右的。