代码随想录算法训练营day18 | leetcode 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

LeetCode 513.找树左下角的值

分析1.0

二叉树的 最底层最左边 节点的值，层序遍历获取最后一层首个节点值，记录每一层的首个节点，当没有下一层时，返回这个节点

class Solution {
    ArrayDeque<TreeNode> queue = new ArrayDeque();
    int res = 0;
    public int findBottomLeftValue(TreeNode root) {
        queue.offer(root);
        return levelOrder(root);
    }

    public int levelOrder(TreeNode p){
        while(!queue.isEmpty()){
            int size = queue.size();
            int cnt = 0;
            res = queue.peek().val;
            // System.out.println("每层第一个节点"+res);
            while(cnt++ < size){
                p = queue.poll();
                if(p.left != null){
                    queue.offer(p.left);
                }
                if(p.right != null){
                    queue.offer(p.right);
                }
            }
        }
        return res;
    }
}

LeetCode 112. 路径总和

分析1.0

先序遍历递归，记录走过节点和，若==targetSum return；否则删除节点值

递归

class Solution {
    int sum = 0;
    ArrayList<Integer> list = new ArrayList();
    public boolean hasPathSum(TreeNode root, int targetSum) {
        if(root == null){
            return false;
        }
        preOrder(root);
        return list.contains(targetSum);
    }
    public void preOrder(TreeNode p){
        sum += p.val;
        if(p.left == null && p.right == null){
            //System.out.println("sum--------"+sum);
            list.add(sum);
            return;
        }
        if(p.left!=null){
            preOrder(p.left);
            sum -= p.left.val;
        }
        if(p.right!=null){
            preOrder(p.right);
            sum -= p.right.val;
        }
    }
}

论递归有返回值时，某路径和为targetSum时，各级递归该如何返回

if (cur->left) { // 左
    count -= cur->left->val; // 递归，处理节点;
    if (traversal(cur->left, count)) return true;
    count += cur->left->val; // 回溯，撤销处理结果
}
if (cur->right) { // 右
    count -= cur->right->val; // 递归，处理节点;
    if (traversal(cur->right, count)) return true;
    count += cur->right->val; // 回溯，撤销处理结果
}
return false;

LeetCode 106.从中序与后序遍历序列构造二叉树

分析1.0

重要理论知识

切割后的左右子树，在前后序两个数组中元素大小是一致的

重要条件：

inorder 和 postorder 都由不同的值组成

后序遍历左右根所以每棵树的后序数组的最后一个节点为树根

从后序数组找到这棵树的树根，用其分割中序数组，使原中序数组变为新的两个中序数组-即两棵子树
子树的大小是一致的，即可利用此特点在后序数组中找到两棵树分别对应的后序遍历序列第一个是左子树第二个是右子树顺序不要弄反
再从后序序列获取新的根节点...
......
直到后序数组只剩一个节点，处理完这个节点就返回

getInIndex(); return 两棵子树在中序数组中的索引

getPostIndex(); return 左右子树的新的根节点索引左边一堆为左子树右边一堆为右子树

失误

没课树都有post in order，当postOrder只剩一个节点时，意味着这棵树只有一个根节点了，让它做父节点合适的儿子

递归前要进行一次判断，这个节点能否满足递归条件

分析2.0

class Solution {
    //int testCount = 1;
    public TreeNode buildTree(int[] inorder, int[] postorder) {

       return  getRootIndex(inorder, postorder, 0, inorder.length - 1, 0, postorder.length-1);
    }

    public TreeNode getRootIndex(int[] inorder, int[] postorder, int inLeftIndex, int inRightIndex,int postLeftIndex,int postRightIndex){
        //System.out.println(testCount++ +"次访问");
        // 只剩一个后序节点了
        if(postRightIndex==postLeftIndex){
            return new TreeNode(postorder[postRightIndex]);
        }
        // 后序根节点在中序中的索引
        int mid = getIndex(inorder, postorder[postRightIndex]);
        // 左子树中序新索引范围
        int leftTreeLeftIndex = inLeftIndex;
        int leftTreeRightIndex = mid-1;
        // 右子树中序新索引范围
        int rightTreeLeftIndex = mid+1;
        int rightTreeRightIndex = inRightIndex;
        /* 找左右子树的后序索引范围
           左子树-右子树-根节点
           左子树 leftTreeRightIndex - leftTreeLeftIndex + 1
           右子树 rightTReeRightIndex - rightTreeLeftIndex + 1        
        */   
        //int leftTreeSize = leftTreeRightIndex - leftTreeLeftIndex + 1;
        int rightTreeSize = rightTreeRightIndex - rightTreeLeftIndex + 1;
        int leftTreePostOrderLeftIndex = postLeftIndex;
        int leftTreePostOrderRightIndex = postRightIndex - rightTreeSize - 1;
        int rightTreePostOrderLeftIndex = postRightIndex - rightTreeSize;
        int rightTreePostOrderRightIndex = postRightIndex - 1;
        TreeNode root = new TreeNode(postorder[postRightIndex]);
        if(leftTreePostOrderRightIndex>=leftTreePostOrderLeftIndex){
            root.left = getRootIndex(inorder, postorder,leftTreeLeftIndex, leftTreeRightIndex,leftTreePostOrderLeftIndex,leftTreePostOrderRightIndex);
        }
        if(rightTreePostOrderRightIndex>=rightTreePostOrderLeftIndex){
            root.right = getRootIndex(inorder, postorder,rightTreeLeftIndex, rightTreeRightIndex,rightTreePostOrderLeftIndex,rightTreePostOrderRightIndex);   
        }
        return root; 
    }

    // 在指定数组中找指定元素的索引
    public int getIndex(int[] arr, int target){
        for(int i = 0; i < arr.length; i++){
            if(arr[i] == target){
                return i;
            }
        }
        return -1;
    }
}