hot100之二叉树下，

hot100之二叉树下，

二叉树的右视图(199)

class Solution {
    List<Integer> res = new ArrayList<>();
    public List<Integer> rightSideView(TreeNode root) {
        dfs(root, 0);
        return res;
    }
    private void dfs(TreeNode node, int depth){
        if (node == null) return;

        if (res.size() == depth){
            res.add(node.val);
        }
        
        dfs(node.right, depth+1);
        dfs(node.left, depth+1);
    }
}

• 分析

因为一层只需要一个节点, 以depth作为限制

二叉树展开为链表(114)

class Solution {
    public void flatten(TreeNode root) {
        if (root == null) return;
        
        flatten(root.left);
        flatten(root.right);
        TreeNode lef = root.left;
        TreeNode rig = root.right;

        root.left = null;
        root.right = lef;
        TreeNode curr = root;
        
        while (curr.right != null){
            curr = curr.right;
        }curr.right = rig;

    }
}

• 分析

将左子树截断放到右端, 再将右子树放在左子树末尾

因为递归原因, 左子树必然为链表

从前序与中序遍历序列构造二叉树(105)

class Solution {
    Map<Integer, Integer> map;
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        int n = preorder.length;
        map = new HashMap<>();
        for (int i = 0; i < n; i++){
            map.put(inorder[i], i);
        }

        return dfs(preorder, 0, n, 0, n);
    }

    private TreeNode dfs(int[] preorder, int lefPre, int rigPre, int lefIn, int rigIn){
        if (lefPre == rigPre) return null;

        int lefSize = map.get(preorder[lefPre]) - lefIn;
        TreeNode lefNode = dfs(preorder, lefPre+1, lefPre+1+lefSize, lefIn, lefIn+lefSize);
        TreeNode rigNode = dfs(preorder, lefPre+1+lefSize, rigPre, lefIn+1+lefSize, rigIn);
        return new TreeNode(preorder[lefPre], lefNode, rigNode);
    }
}

• 分析

以前序遍历取根, 切分中序遍历的左右子树作构造

路径总和(437)

class Solution {
    Map<Long, Integer> map = new HashMap<>();
    int res;
    public int pathSum(TreeNode root, int targetSum) {
        map.put(0L,1);
        dfs(root, targetSum, 0L);
        return res;
    }

    private void dfs(TreeNode node, int targetSum, long subSum){
        if (node == null) return;
        subSum += node.val;
        if (map.containsKey(subSum - targetSum)) res +=map.get(subSum - targetSum);
        map.put(subSum, map.getOrDefault(subSum, 0)+1);
        dfs(node.left, targetSum, subSum);
        dfs(node.right, targetSum, subSum);
        map.put(subSum, map.get(subSum)-1);
    }
}

• 分析

又是一个前缀和题目

二叉树的最近公共祖先(236)

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || p == root || q == root) return root;

        TreeNode lefNode = lowestCommonAncestor(root.left, p, q);
        TreeNode rigNode = lowestCommonAncestor(root.right, p, q);

        if (lefNode == null && rigNode == null) return null;
        if (lefNode != null && rigNode != null) return root;

        return lefNode != null ? lefNode : rigNode;

    }
}

• 分析

找到子节点向上return, 两节点未相遇则继续向上return

• 感悟

二叉树题目要注意需要返回给父节点的元素

二叉树中的最大路径和(124)

class Solution {
    int res = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        dfs(root);
        return res;
    }
    private int dfs(TreeNode node){
        if (node == null) return 0;

        int lefMax = Math.max(dfs(node.left), 0);
        int rigMax = Math.max(dfs(node.right), 0);

        res = Math.max(res, lefMax+ node.val + rigMax);
        return Math.max(lefMax, rigMax) + node.val;
    }
}

• 分析

以单向链作返回父节点元素, 取 左右链+节点 与记录的最大值作对比