leetCode热题100-47、48题

47、路径总和

给定一个二叉树的根节点 root ，和一个整数 targetSum ，求该二叉树里节点值之和等于 targetSum 的 路径 的数目。

路径 不需要从根节点开始，也不需要在叶子节点结束，但是路径方向必须是向下的（只能从父节点到子节点）。

/**
 * 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 pathSum(TreeNode root, int targetSum) {
        Map<Long,Integer> prefix = new HashMap<Long,Integer>();
        prefix.put(0L,1);
        return dfs(root,prefix,0,targetSum);       
    }

    public int dfs(TreeNode root,Map<Long,Integer> prefix,long curr,int targetSum) {
        if (root == null) {
            return 0;
        }

        int ret = 0;
        curr += root.val;

        ret = prefix.getOrDefault(curr - targetSum,0);
        prefix.put(curr,prefix.getOrDefault(curr,0) + 1);
        ret += dfs(root.left,prefix,curr,targetSum);
        ret += dfs(root.right,prefix,curr,targetSum);
        prefix.put(curr,prefix.getOrDefault(curr,0) - 1);

        return ret;
    }
}

48、找树中两个指定节点的最近公共祖先

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {

    private TreeNode ans;

    public Solution() {
        this.ans = null;
    }

    private boolean dfs(TreeNode root,TreeNode p,TreeNode q) {
        if (root == null) return false;
        boolean lson = dfs(root.left,p,q);
        boolean rson = dfs(root.right,p,q);
        if ((lson && rson) || ((root.val = p.val || root.val == q.val) && (lson || rson))) {
            ans = root;
        }
        return lson || rson || (root.val == p.val || root.val == q.val);
    }

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        this.dfs(root,p,q);
        return this.ans;   
    }
}